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Index1. Theory2. Short Revision3. Exercise (1 to 5)4. Assertion & Reason5. Que. from Compt. Exams
Subject : Mathematics
Topic: Complex Number
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1. The complex number systemThere is no real number x which satisfies the polynomial equation x2 + 1 = 0. To permit solutions of thisand similar equations, the set of complex numbers is introduced.We can consider a complex number as having the form a + bi where a and b are real number and i,which is called the imaginary unit, has the property that i2 = – 1.It is denoted by z i.e. z = a + ib. ‘a’ is called as real part of z which is denoted by (Re z) and ‘b’ is calledas imaginary part of z which is denoted by (Im z).Any complex number is :(i) Purely real, if b = 0 ; (ii) Purely imaginary, if a = 0(iii) Imaginary, if b ≠ 0.
NOTE : (a) The set R of real numbers is a proper subset of the Complex Numbers. Hence the completenumber system is N ⊂ W ⊂ I ⊂ Q ⊂ R ⊂ C.(b) Zero is purely real as well as purely imaginary but not imaginary.(c) i = −1 is called the imaginary unit.
Also i² = − 1; i3 = − i ; i4 = 1 etc.
(d) a b = a b only if atleast one of a or b is non - negative.(e) is z = a + ib, then a – ib is called complex conjugate of z and written as z = a – ib
Self Practice Problems1. Write the following as complex number
(i) 16− (ii) x , (x > 0)
(iii) –b + ac4− , (a, c> 0)
Ans. (i) 0 + i 16 (ii) x + 0i (iii) –b + i ac42. Write the following as complex number
(i) x (x < 0) (ii) roots of x2 – (2 cosθ)x + 1 = 02. Algebraic Operations:
Fundamental operations with complex numbersIn performing operations with complex numbers we can proceed as in the algebra of real numbers,replacing i2 by – 1 when it occurs.1. Addition (a + bi) + (c + di) = a + bi + c + di = (a + c) + (b + d) i2. Subtraction (a + bi) – c + di) = a + bi – c – di = (a – c) + (b – d) i3. Multiplication (a + bi) (c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad+ bc)i
4. Division dicbia
++
= dicbia
++
. dicbic
−−
= 222
2
idcbdibciadiac
−−+−
= 22 dci)adbc(bdac
−−++
= 22 dcbdac
++
+ idcadbc
22 +−
Inequalities in complex numbers are not defined. There is no validity if we say that complex number ispositive or negative.e.g. z > 0, 4 + 2i < 2 + 4 i are meaningless.In real numbers if a2 + b2 = 0 then a = 0 = b however in complex numbers,z1
2 + z22 = 0 does not imply z1 = z2 = 0.
Example : Find multiplicative inverse of 3 + 2i.Solution Let z be the multiplicative inverse of 3 + 2i. then
⇒ z . (3 + 2i) = 1
⇒ z = i231+ = ( ) ( )i23i23
i23−+
−
⇒ z = 133
– 132
i
− i
132
133
Ans.Self Practice Problem1. Simplify in+100 + in+50 + in+48 + in+46 , n ∈ Ι .
Ans. 0
3. Equality In Complex Number:Two complex numbers z1 = a1 + ib1 & z2 = a2 + ib2 are equal if and only if their real and imaginary partsare equal respectivelyi.e. z1 = z2 ⇒ Re(z1) = Re(z2) and Ιm (z1) = Ιm (z2).
Complex Numbers
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omExample: Find the value of x and y for which (2 + 3i) x2 – (3 – 2i) y = 2x – 3y + 5i where x, y ∈ R.Solution (z + 3i)x2 – (3 – 2i)y = 2x – 3y + 5i
⇒ 2x2 – 3y = 2x – 3y⇒ x2 – x = 0⇒ x = 0, 1 and 3x2 + 2y = 5
⇒ if x = 0,y = 25
and if x = 1, y = 1
∴ x = 0, y = 25
and x = 1, y = 1
are two solutions of the given equation which can also be represented as
25,0 & (1, 1)
25,0 , (1, 1) Ans.
Example: Find the value of expression x4 – 4x3 + 3x2 – 2x + 1 when x = 1 + i is a factor of expression.Solution. x = 1 + i
⇒ x – 1 = i⇒ (x – 1)2 = –1⇒ x2 – 2x + 2 = 0Now x4 – 4x3 + 3x2 – 2x + 1
= (x2 – 2x + 2) (x2 – 3x – 3) – 4x + 7∴ when x = 1 + i i.e. x2 – 2x + 2 = 0
x4 – 4x3 + 3x2 – 2x + 1 = 0 – 4 (1 + i) + 7= –4 + 7 – 4i= 3 – 4i Ans.
Example: Solve for z if z2 + |z| = 0Solution. Let z= x + iy
⇒ (x + iy)2 + 22 yx + = 0
⇒ x2 – y2 + 22 yx + = 0 and 2xy = 0⇒ x = 0 or y = 0when x = 0 – y2 + | y | = 0⇒ y = 0, 1, –1⇒ z = 0, i, –iwhen y = 0 x2 + | x | = 0⇒ x = 0 ⇒ z = 0 Ans. z = 0, z = i, z = – i
Example: Find square root of 9 + 40iSolution. Let (x + iy)2 = 9 + 40i
∴ x2 – y2 = 9 ...............(i)and xy = 20 ...............(ii)squing (i) and adding with 4 times the square of (ii)we get x4 + y4 – 2x2 y2 + 4x2 y2 = 81 + 1600⇒ (x2 + y2)2 = 168⇒ x2 + y2 = 4 ...............(iii)from (i) + (iii) we get x2 = 25 ⇒ x = ± 5
and y = 16 ⇒ y = ± 4from equation (ii) we can see thatx & y are of same sign∴ x + iy = +(5 + 4i) or = (5 + 4i)∴ sq. roots of a + 40i = ± (5 + 4i) Ans. ± (5 + 4i)
Self Practice Problem
1. Solve for z : z = i z2 Ans. ± 23
– 21
i, 0, i
4. Representation Of A Complex Number:(a) Cartesian Form (Geometric Representation) :
Every complex number z = x + i y can be represented by a point on the Cartesian planeknown as complex plane (Argand diagram) by the ordered pair (x, y).
Length OP is called modulus of the complex number which is denoted by z & θ is called theargument or amplitude.
z = x y2 2+ & θ = tan−1yx (angle made by OP with positive x−axis)
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omNOTE : (i) Argument of a complex number is a many valued function. If θ is the argument of a complex
number then 2 nπ + θ; n ∈ I will also be the argument of that complex number. Any two arguments ofa complex number differ by 2nπ.(ii) The unique value of θ such that − π < θ ≤ π is called the principal value of the argument.
Unless otherwise stated, amp z implies principal value of the argument.(iii) By specifying the modulus & argument a complex number is defined completely. For the complex
number 0 + 0 i the argument is not defined and this is the only complex number which is onlygiven by its modulus.
(b) Trignometric/Polar Representation :z = r (cos θ + i sin θ) where z = r; arg z = θ ; z = r (cos θ − i sin θ)
NOTE : cos θ + i sin θ is also written as CiS θ or ei θ.
Also cos x =2ee ixix −+ & sin x =
2ee ixix −− are known as Euler's identities.
(c) Euler's Representation :z = reiθ; z = r; arg z = θ; z = re− iθ
(d) Vectorial Representation :Every complex number can be considered as if it is the position vector of a point. If the point
P represents the complex number z then, OP→
= z & OP→
= z.Example: Express the complex number z = – 1 + 2 i in polar form.
Solution. z = –1 + i 2
| z | = ( )22 2)1( +− = 21+ = 3
Arg z = π – tan–1
12
= π – tan–1 2 = θ (say)
∴ z = 3 (cos θ + i sin θ ) where θ = π – tan–1 2Self Practice Problems
1. Find the principal argument and |z|
z = i2
)i9(1−
+−
Ans. – tan–1 1117
, 582
2. Find the |z| and principal argument of the complex number z = 6(cos 310º – i sin 310°)Ans. 6, 50°
5. Modulus of a Complex Number :If z = a + ib, then it's modulus is denoted and defined by |z| = 22 ba + . Infact |z| is the distanceof z from origin. Hence |z1 – z2| is the distance between the points represented by z1 and z2.
Properties of modulus
(i) |z1z2| = |z1| . |z2| (ii)2
1
zz
= 2
1
zz
(provided z2 ≠ 0)
(iii) |z1 + z2| ≤ |z1| + |z2| (iv) |z1 – z2| ≥ ||z1| – |z2||
(Equality in (iii) and (iv) holds if and only if origin, z1 and z2 are collinear with z1 and z2 on the same sideof origin).
Example: If |z – 5 – 7i| = 9, then find the greatest and least values of |z – 2 – 3i|.Solution. We have 9 = |z – (5 + 7i)| = distance between z and 5 + 7i.
Thus locus of z is the circle of radius 9 and centre at 5 + 7i. For such a z (on the circle), wehave to find its greatest and least distance as from 2 + 3i, which obviously 14 and 4.
Example: Find the minimum value of |1 + z| + |1 – z|.Solution |1 + z| + |1 – z| ≥ |1 + z + 1 – z| (triangle inequality)
⇒ |1 + z | + |1 – z| ≥ 2∴ minimum value of (|1 + z| + |1 – z|) = 2Geometrically |z + 1| + |1 – 2| = |z + 1| + |z – 1| which represents sum of distances of z from1 and – 1it can be seen easily that minimu (PA + PB) = AB = 2
Ans.
π+π n
81
4/1 e2
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omExample: z
2z − = 1 then find the maximum and minimum value of |z|
Solution. z2z − = 1 z
2|z| − ≤ 22z − ≤ | z | + z
2−
Let | z | = r
⇒ r2r − ≤ 1 ≤ r +
r2
r + r2
≥ 1 ⇒ r ∈ R+ ..............(i)
and r2r − ≤ 1 ⇒ –1 ≤ r –
r2
≤ 1⇒ r ∈ (1, 2) ..............(ii)∴ from (i) and (ii) r ∈ (1, 2)Ans. r ∈ (1, 2)
Self Practice Problem
1. |z – 3| < 1 and |z – 4i| > M then find the positive real value of M for which these exist at least onecomplex number z satisfy both the equation.Ans. M ∈ (0, 6)
6. Agrument of a Complex Number :
Argument of a non-zero complex number P(z) is denoted and defined by arg(z) = angle which OPmakes with the positive direction of real axis.If OP = |z| = r and arg(z) = θ, then obviously z = r(cosθ + isinθ), called the polar form of z. In whatfollows, 'argument of z' would mean principal argument of z(i.e. argument lying in (– π, π] unless thecontext requires otherwise. Thus argument of a complex number z = a + ib = r(cosθ + isinθ) is the valueof θ satisfying rcosθ = a and rsinθ = b.
Thus the argument of z = θ, π – θ, – π + θ, – θ, θ = tan–1 a
b , according as z = a + ib lies in Ι, ΙΙ , ΙΙΙor ΙVth quadrant.
Properties of arguments(i) arg(z1z2) = arg(z1) + arg(z2) + 2mπ for some integer m.(ii) arg(z1/z2) = arg (z1) – arg(z2) + 2mπ for some integer m.(iii) arg (z2) = 2arg(z) + 2mπ for some integer m.(iv) arg(z) = 0 ⇔ z is real, for any complex number z ≠ 0(v) arg(z) = ± π/2 ⇔ z is purely imaginary, for any complex number z ≠ 0(vi) arg(z2 – z1) = angle of the line segment
P′Q′ || PQ, where P′ lies on real axis, with the real axis.
Example: Solve for z, which satisfy Arg (z – 3 – 2i) = 6π
and Arg (z – 3 – 4i) = 32π
.Solution From the figure, it is clear that there is no z, which satisfy both ray
Example: Sketch the region given by(i) Arg (z – 1 – i) ≥ π/3(ii) |z| = ≤ 5 & Arg (z – i – 1) > π/3
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Solution (i) (ii)
Self Practice Problems
1. Sketch the region given by(i) |Arg (z – i – 2)| < π/4 (ii) Arg (z + 1 – i) ≤ π/6
2. Consider the region |z – 15i| ≤ 10. Find the point in the region which has(i) max |z| (ii) min |z|(iii) max arg (z) (iv) min arg (z)
7. Conjugate of a complex Number
Conjugate of a complex number z = a + b is denoted and defined by z = a – ib.In a complex number if we replace i by – i, we get conjugate of the complex number. z is the mirrorimage of z about real axis on Argand's Plane.
Properties of conjugate
(i) |z| = | z | (ii) z z = |z|2
(iii) )zz( 21 + = )z( 1 + )z( 2 (iv) )zz( 21 − = )z( 1 – )z( 2
(v) )zz( 21 = 1z 2z (vi)
2
1zz
= )z()z(
2
1 (z2 ≠ 0)
(vii) |z1 + z2|2 = (z1 + z2) )zz( 21 + = |z1|2 + |z2|2 + z1 2z + 1z z2
(viii) )z( 1 = z (ix) If w = f(z), then w = f( z )(x) arg(z) + arg( z ) = 0
Example: If 1z1z
+−
is purely imaginary, then prove that | z | = 1
Solution. Re
+−
1z1z
= 0
⇒1z1z
+−
+
+−
1z1z
= 0 ⇒1z1z
+−
+ 1z1z
+−
= 0
⇒ z z – z + z – 1 + z z – z + z – 1 = 0⇒ z z = 1 ⇒ | z |2 = 1⇒ | z | = 1 Hence proved
Self Practice Problem
1. If 21
21
zz2z2z
−−
is unmodulus and z2 is not unimodulus then find |z1|.
Ans. |z1| = 28. Rotation theorem(i) If P(z1) and Q(zz) are two complex numbers such that |z1| = |z2|, then z2 = z1 eiθ where θ = ∠ POQ(ii) If P(z1), Q(z2) and R(z3) are three complex numbers and ∠ PQR = θ, then
−−
21
23
zzzz
= 21
23
zzzz
−−
eiθ
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7 of 3
8
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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dent
s ava
ilabl
e at
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site:
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ww
.teko
clas
ses.c
om(iii) If P(z1), Q(z2), R(z3) and S(z4) are four complex numbers and ∠ STQ = θ, then
21
23
zzzz
−−
= 21
43
zzzz
−−
eiθ
Example: If arg
+−
iz1z
= 3π
then interrupter the locus.
Solution arg
+−
iz1z
= 3π
⇒ arg
−−−
z1z1
= 3π
Here arg
−−−
z1z1
represents the angle between lines joining –1 and z and 1 + z. As this angleis constant, the locus of z will be a of a circle segment. (angle in a segment is count). It can be
seen that locus is not the complete side as in the major are arg
−−−
z1z1
will be equal to – 32π
.Now try to geometrically find out radius and centre of this circle.
centre ≡
31,0 Radius ≡ 3
2Ans.
Example: If A(z + 3i) and B(3 + 4i) are two vertices of a square ABCD (take in anticlock wise order) thenfind C and D.
Solution. Let affix of C and D are z3 + z4 respectivelyConsidering ∠ DAB = 90º + AD = AB
we getAD
)i32(z4 +− =
AB)i32()i43( +−+
e 2i π
⇒ z4 – (2 + 3i) = (1 + i) i⇒ Z4 = 2 + 3i+ i – 1 = 1 + zi
and CB)i43(z3 +−
= AB
)i43()i3z( −−+e –
2i π
⇒ z3 = 3 + 4i – (1 + i) (–i)z3 = 3 + 4i + i – 1 = z + 5i
Self Practice Problems
1. z1, z2, z3, z4 are the vertices of a square taken in anticlockwise order then prove that2z2 = (1 + i) z1 + (1 – i) z3Ans. (1 + i) z1 + (1 – i)z3
2. Check that z1z2 and z3z4 are parallel or, notwhere, z1 = 1 + i z3 = 4 + 2i
z2 = 2 – i z4 = 1 – iAns. Hence, z1z2 and z3z4 are not parallel.
3. P is a point on the argand diagram on the circle with OP as diameter “two point Q and R are taken suchthat ∠ POQ = ∠ QORIf O is the origin and P, Q, R are represented by complex z1, z2, z3 respectively then show that
z22 cos 2θ = z1z3cos2θ
Ans. z1z3 cos2θ
9. Demoivre’s Theorem:Case ΙΙΙΙΙStatement :If n is any integer then(i) (cos θ + i sin θ )n = cos nθ + i sin nθ(ii) (cos θ1 + i sin θ1) (cos θ2) + i sin θ2) (cosθ3 + i sin θ2) (cos θ3 + i sin θ3) .....(cos θn + i sin θn)
= cos (θ1 + θ2 + θ3 + ......... θn) + i sin (θ1 + θ2 + θ3 + ....... + θn)Case ΙΙΙΙΙΙΙΙΙΙStatement : If p, q ∈ Z and q ≠ 0 then
(cos θ + i sin θ)p/q = cos
θ+πq
pk2 + i sin
θ+πq
pk2
where k = 0, 1, 2, 3, ......, q – 1
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8 of 3
8
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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s ava
ilabl
e at
web
site:
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tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
omNOTE : Continued product of the roots of a complex quantity should be determined using theory of equations.
10. Cube Root Of Unity :
(i) The cube roots of unity are 1, − +1 32
i , − −1 32
i .(ii) If ω is one of the imaginary cube roots of unity then 1 + ω + ω² = 0. In general 1 + ωr + ω2r = 0;
where r ∈ I but is not the multiple of 3.(iii) In polar form the cube roots of unity are :
cos 0 + i sin 0; cos23π
+ i sin23π
, cos43π
+ i sin 43π
(iv) The three cube roots of unity when plotted on the argand plane constitute the verties of anequilateral triangle.
(v) The following factorisation should be remembered :(a, b, c ∈ R & ω is the cube root of unity)a3 − b3 = (a − b) (a − ωb) (a − ω²b) ; x2 + x + 1 = (x − ω) (x − ω2) ;a3 + b3 = (a + b) (a + ωb) (a + ω2b) ; a2 + ab + b2 = (a – bw) (a – bw2)a3 + b3 + c3 − 3abc = (a + b + c) (a + ωb + ω²c) (a + ω²b + ωc)
Example: Find the value of ω192 + ω194
Solution. ω192 + ω194
= 1 + ω2 = – ωAns. – ω
Example: If 1, ω, ω2 are cube roots of unity prove(i) (1 – ω + ω2) (1 + ω – ω2) = 4(ii) (1 – ω + ω2)5 + (1 + ω – ω2)5 = 32(iii) (1 – ω) (1 – ω2) (1 – ω4) (1 – ω8) = 9(iv) (1 – ω + ω2) (1 – ω2 + ω4) (1 – ω4 + ω8) .......... to 2n factors = 22n
Solution. (i) (1 – ω + ω2) (1 + ω – ω2)= ( – 2ω) ( – 2ω2)= 4
Self Practice Problem
1. Find ∑=
ω+ω+10
0r
r2r )1(
Ans. 1211. nth Roots of Unity :
If 1, α1, α2, α3..... αn − 1 are the n, nth root of unity then :
(i) They are in G.P. with common ratio ei(2π/n) &
(ii) 1p + α 1p + α 2
p +.... +α np
−1 = 0 if p is not an integral multiple of n = n if p is an integral multiple of n
(iii) (1 − α1) (1 − α2)...... (1 − αn − 1) = n &(1 + α1) (1 + α2)....... (1 + αn − 1) = 0 if n is even and 1 if n is odd.
(iv) 1. α1. α2. α3......... αn − 1 = 1 or −1 according as n is odd or even.Example: Find the roots of the equation z6 + 64 = 0 where real part is positive.Solution. z6 = – 64
z6 = z6 . e+ i(2n + 1)π x ∈ z
⇒ z = z 6)1n2(i
eπ+
∴ z = 2 6i
eπ
, 2 2i
eπ
, z 2i
eπ
, z 65i
eπ
= 67i
eπ
, z 23i
eπ
, z 211i
eπ
∴ roots with +ve real part are = 6i
eπ
+ 611i
eπ
π−
6i
e2 Ans.
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9 of 3
8
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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Pack
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s ava
ilabl
e at
web
site:
ww
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tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
omExample: Find the value ∑
=
π−π6
1k 7k2cos
7k2sin
Solution. ∑=
π6
1k 7k2sin – ∑
=
π6
1k 7k2cos
= ∑=
π6
0k 7k2sin – ∑
=
π6
0k 7k2cos + 1
= ∑=
6
0k (Sum of imaginary part of seven seventh roots of unity)
– ∑=
6
0k (Sum of real part of seven seventh roots of unity) + 1
= 0 – 0 + 1 = 1i Ans.
Self Practice Problems
1. Resolve z7 – 1 into linear and quadratic factor with real coefficient.
Ans. (z – 1)
+π− 1z
72cos2z2
.
+π− 1z
74cos2z2
.
+π− 1z
76cos2z2
2. Find the value of cos 72π
+ cos 74π
+ cos 76π
.
Ans. – 21
12. The Sum Of The Following Series Should Be Remembered :
(i) cos θ + cos 2 θ + cos 3 θ +..... + cos n θ =( )( )
sin /sin /
nθθ
22 cos
n +
12 θ.
(ii) sin θ + sin 2 θ + sin 3 θ +..... + sin n θ =( )( )
sin /sin /
nθθ
22 sin
n +
12 θ.
NOTE : If θ = (2π/n) then the sum of the above series vanishes.
13. Logarithm Of A Complex Quantity :
(i) Loge (α + i β) = 12
Loge (α² + β²) + i 2 1n π βα
+
−tan where n ∈ Ι.
(ii) ii represents a set of positive real numbers given by en− +
22
π π
, n ∈ Ι.
Example: Find the value of
(i) log (1 + 3 i) Ans. log2 + i(2nπ + 3π
)(ii) log(–1) Ans. iπ(iii) zi Ans. cos(ln2) + i sin(ln2) = ei(ln2)
(iv) ii Ans. 2).1n4(
eπ+−
(v) |(1 + i)i | Ans. 4).1n8(
eπ+−
(vi) arg ((1 + i)i) Ans.21
n(2).
Solution. (i) log (1 + 3 i) = log
π+π n2
3i
e2
= log 2 + i
π+π n2
3(iii) 2i = ei n 2 = cos ( n 2) cos ( n 2) + i sin ( n 2) ]
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10 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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e at
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site:
ww
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tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
omSelf Practice Problem
1. Find the real part of cos (1 + i)
Ans.ei2e1 2−
14. Geometrical Properties :Distance formula :If z1 and z2 are affixies of the two points ↓ P and Q respectively then distance between P + Q is givenby |z1 – z2|.Section formulaIf z1 and z2 are affixes of the two points P and Q respectively and point C devides the line joining P andQ internally in the ratio m : n then affix z of C is given by
z = nmnzmz 12
++
If C devides PQ in the ratio m : n externally then
z = nmnzmz 12
−−
(b) If a, b, c are three real numbers such that az1 + bz2 + cz3 = 0 ; where a + b + c = 0 and a,b,care not all simultaneously zero, then the complex numbers z1, z2 & z3 are collinear.
(1) If the vertices A, B, C of a ∆ represent the complex nos. z1, z2, z3 respectively anda, b, c are the length of sides then,
(i) Centroid of the ∆ ABC =z z z1 2 3
3+ +
:(ii) Orthocentre of the ∆ ABC =
( ) ( ) ( )CseccBsecbAseca
zCsecczBsecbzAseca 321++
++ or CtanBtanAtan
CtanzBtanzAtanz 321++++
(iii) Incentre of the ∆ ABC = (az1 + bz2 + cz3) ÷ (a + b + c).
(iv) Circumcentre of the ∆ ABC = :(Z1 sin 2A + Z2 sin 2B + Z3 sin 2C) ÷ (sin 2A + sin 2B + sin 2C).
(2) amp(z) = θ is a ray emanating from the origin inclined at an angle θ to the x− axis.(3) z − a = z − b is the perpendicular bisector of the line joining a to b.(4) The equation of a line joining z1 & z2 is given by, z = z1 + t (z1 − z2) where t is a real parameter.(5) z = z1 (1 + it) where t is a real parameter is a line through the point z1 & perpendicular to the
line joining z1 to the origin.(6) The equation of a line passing through z1 & z2 can be expressed in the determinant form as
z zz zz z
111
1 1
2 2
= 0. This is also the condition for three complex numbers to be collinear. The above
equation on manipulating, takes the form α αz z r+ + = 0 where r is real and α is a non zerocomplex constant.
NOTE : If we replace z by zeiθ and z by θi–ez then we get equation of a straight line which. Passes through thefoot of the perpendicular from origin to given straight line and makes an angle θ with the given straightlline.
(7) The equation of circle having centre z0 & radius ρ is : z − z0 = ρ or z z − z0 z − z0 z + z0 z0 − ρ² = 0 which is of the form
z z z z+ +α α + k = 0, k is real. Centre is − α & radius = αα − k .
Circle will be real if αα − k ≥ 0..
(8) The equation of the circle described on the line segment joining z1 & z2 as diameter is
arg z zz z
−−
2
1
= ± π2
or (z − z1) ( z − z 2) + (z − z2) ( z − z 1) = 0.
(9) Condition for four given points z1, z2, z3 & z4 to be concyclic is the numberz zz z
z zz z
3 1
3 2
4 2
4 1
−−
−−
. should be real. Hence the equation of a circle through 3 non collinear
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11 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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e at
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clas
ses.c
ompoints z1, z2 & z3 can be taken as
( ) ( )( ) ( )z z z zz z z z
− −− −
2 3 1
1 3 2 is real
⇒ ( ) ( )( ) ( )z z z zz z z z
− −− −
2 3 1
1 3 2
=( ) ( )( ) ( )z z z zz z z z
− −− −
2 3 1
1 3 2.
(10) Arg
−−
2
1zzzz
= θ represent (i) a line segment if θ = π
(ii) Pair of ray if θ = 0 (iii) a part of circle, if 0 < θ < π.
(11) Area of triangle formed by the points z1, z2 & z3 is 1zz1zz1zz
i41
33
22
11
(12) Perpendicular distance of a point z0 from the line 0rzz =+α+α is ||2|rzz| 00
α+α+α
(13) (i) Complex slope of a line 0rzz =+α+α is ω = – αα
.
(ii) Complex slope of a line joining by the points z1 & z2 is ω = 21
21
zzzz
−−
(iii) Complex slope of a line making θ angle with real axis = e2iθ
(14) ω1 & ω2 are the compelx slopes of two lines.(i) If lines are parallel then ω1 = ω2(ii) If lines are perpendicular then ω1 + ω2 = 0
(15) If |z – z1| + |z – z2| = K > |z1 – z2| then locus of z is an ellipse whose focii are z1 & z2
(16) If |z – z0| = ||2rzz
α+α+α
then locus of z is parabola whose focus is z0 and directrix is the
line 0zα + 0zα + r = 0
(17) If 2
1zzzz
−−
= k ≠ 1, 0, then locus of z is circle.
(18) If z – z1 – z – z2 = K < z1 – z2 then locus of z is a hyperbola, whose focii arez1 & z2.
Match the following columns :Column - ΙΙΙΙΙ Column - ΙΙΙΙΙΙΙΙΙΙ
(i) If | z – 3+2i | – | z + i | = 0, (i) circlethen locus of z represents ..........
(ii) If arg
+−
1z1z
= 4π
, (ii) Straight linethen locus of z represents...
(iii) if | z – 8 – 2i | + | z – 5 – 6i | = 5 (iii) Ellipsethen locus of z represents .......
(iv) If arg
−++−
i52zi43z
= 65π
, (iv) Hyperbola
then locus of z represents .......
(v) If | z – 1 | + | z + i | = 10 (v) Major Arcthen locus of z represents ........
(vi) | z – 3 + i | – | z + 2 – i | = 1 (vi) Minor arcthen locus of z represents .....
(vii) | z – 3i | = 25 (vii) Perpendicular bisector of a line segment
(viii) arg
++−iz
i53z = π (viii) Line segment
Ans. ΙΙΙΙΙ (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)ΙΙΙΙΙΙΙΙΙΙ (vii) (v) (viii) (vi) (iii) (iv) (i) (ii)
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om15. (a) Reflection points for a straight line :
Two given points P & Q are the reflection points for a given straight line if the given line is theright bisector of the segment PQ. Note that the two points denoted by the complexnumbers z1 & z2 will be the reflection points for the straight line α αz z r+ + = 0 if and only if;α αz z r1 2 0+ + = , where r is real and α is non zero complex constant.
(b) Inverse points w.r.t. a circle :Two points P & Q are said to be inverse w.r.t. a circle with centre 'O' and radius ρ, if:(i) the point O, P, Q are collinear and P, Q are on the same side of O.(ii) OP. OQ = ρ2.
Note : that the two points z1 & z2 will be the inverse points w.r.t. the circle z z z z r+ + + =α α 0 if and onlyif z z z z r1 2 1 2 0+ + + =α α .
16. Ptolemy’s Theorem:It states that the product of the lengths of the diagonals of a convex quadrilateral inscribed in a circleis equal to the sum of the products of lengths of the two pairs of its opposite sides.i.e. z1 − z3 z2 − z4 = z1 − z2 z3 − z4 + z1 − z4 z2 − z3.
Example: If cos α + cos β + cos γ = 0 and also sin α + sin β + sin γ = 0, then prove that(i) cos 2α + cos2β + cos2γ = sin 2α + sin 2β + sin 2γ = 0(ii) sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)(iii) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
Solution. Let z1 = cos α + i sin α, z2 = cos β + i sin β,z3 = cosγ + i sin γ.
∴ z1 + z2 + z3 = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)= 0 + i . 0 = 0 (1)
(i) Also1z
1 = (cos α + i sin α)–1 = cos α – i sin α
2z1
= cos β – i sin β, 3z1
– cos γ – sin γ
∴1z
1 +
2z1
+ 3z1
= (cos α + cos β + cos γ) – i (sin α + sin β + sin γ) (2) = 0 – i . 0 = 0
Now z12 + z2
2 + z33 = (z1 + z2 + z3)2 – 2 (z1z2 + z2z3 + z3z1 )
= 0 – 2z1z2z3
++
213 z1
z1
z1
= 0 – 2z1 z2 z3. 0 = 0, using (1) and (2)or (cos α + i sin α)2 + (cos β + i sin β)2 + (cos γ + i sin γ)2 = 0or cos 2α + i sin 2α)2 + cos 2β + i sin 2β + cos 2γ + i sin 2γ = 0 + i.0Equation real and imaginary parts on both sides, cos 2α + cos 2β + cos 2γ = 0 andsin 2α + sin 2β + sin 2γ = 0
(ii) z13 + z2
3 + z33 = (z1 + z2)3 – 3z1z2(z1 + z2) + z3
3
= (–z3)3 – 3z1z2 (– z3) + z33, using (1)
= 3z1z2z3∴ (cos α + i sin α)3 + (cos β + i sin β)3 + (cos γ + i sin γ)3
= 3 (cos α + i sin α) (cos β + i sin β) (cos γ + i sin γ)or cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ
= 3{cos(α + β + γ) + i sin (α + β + γ)Equation imaginary parts on both sides, sin 3α + sin 3β + sin 3γ = 3 sin (α + β + γ)
Alternative methodLet C ≡ cos α + cos β + cos γ = 0
S ≡ sin α + sin β + sin γ = 0C + iS = eiα + eiβ + eiγ = 0 (1)C – iS = e–iα + e–iβ + e–iγ = 0 (2)
From (1) ⇒ (e–iα)2 + (e–iβ)2 + (e–iγ)2 = (eiα) (eiβ) + (eiβ) (eiγ) + (eiγ) (eiα) ⇒ ei2α + ei2β + ei2γ = eiα eiβ eiγ (e–2γ + e–iα + eiβ) ⇒ ei(2α) + ei2β + ei2γ = 0 (from 2)
Comparing the real and imaginary parts wecos 2α + cos 2β + cos 2γ – sin 2α + sin 2β + sin 2γ = 0Also from (1) (eiα)3 + (eiβ)3 + (eiγ)3 = 3eiα eiβ eiγ
⇒ ei3α + ei3β + ei3γ = 3ei(α+β+γ)
Comparing the real and imaginary parts we obtain the results.
Example: If z1 and z2 are two complex numbers and c > 0, then prove that
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13 of
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om|z1 + z2|2 ≤ (I + C) |z1|2 + (I +C–1) |z2|2
Solution. We have to prove :|z1 + z2|2 ≤ (1 + c) |z1|2 + (1 + c–1) |z2|2
i.e. |z1|3 + |z2|2 + z1 z 2 + z 2z2 ≤ (1 + c) |z1|2 + (1 +c–1) |z2|3
or z1 z 2 + z 2z2 ≤ c|z1|2 + c–1|z2|2 or c|z1|2 + c1
|z2|2 – z1 z 2 – z 2 z2 ≥ 0
(using Re (z1 z 2) ≤ |z1 z 2|)
or2
21 |z|c1zc
− ≥ 0 which is always true.
Example: If θ, ∈ [π/6, π/3], i = 1, 2, 3, 4, 5, and z4 cos θ1 + z3 cos θ2 + z3 cos θ3. + z cos θ4 + cosθ5 = 32 ,
then show that |z| > 43
Solution. Given thatcosθ1 . z4 + cosθ2 . z3 + cosθ3 . z2 + cosθ4 . z + cosθ5 = 2√3
or |cosθ1 . z4 + cosθ2 . z3 + cosθ3 . z2 + cosθ4 . z + cosθ5| = 2√32√3 ≤ |cosθ1 . z4 | + |cosθ2 . z3 | + |cosθ3 . z2 | + cosθ4 . z| + |cosθ5 |
∵ θi ∈ [π/6, π/3]
∴21
≤ cosθi ≤ 23
32 ≤ 23
|z|4 + 23
|z|3 + 23
|z|2 + 23
|z| + 23
3 ≤ |z|4 + |z|3 + |z|2 + |z|3 < |z| + |z|2 + |z|3 + |z|4 +|z|5 + .........∞
3 < |z|1|z|
− 3 – e |z| < |z|
4|z| > 3 ∴ |z| > 43
Example: Two different non parallel lines cut the circle |z| = r in point a, b, c, d respectively. Prove that
these lines meet in the point z given by z = 1111
1111
dcbadcba
−−−−
−−−−
−−−+
Solution. Since point P, A, B are collinear
∴1bb1aa1zz
= 0 ⇒ z ( )ba − – z (a – b) + ( )baba − = 0 (i)
Similarlym, since points P, C, D are collinear∴ z ( )ba − (c – d) – z ( )dc − (a – b) = ( )dcdc − (a – b) – ( )baba − (c – d) (iii)
∵ zz = r2 = k (say) ∴ a = ak
, b = bk
, c = ck
etc.From equation (iii) we get
z
−
bk
ak
(c – d) – z
−
dk
ck
(a – b) =
−
ckd
dck
(a – b) –
−
abk
bak
(c – d)
∴ z = 1111
1111
dcbadcba
−−−−
−−−−
−−−+
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ShorShorShorShorShort Rt Rt Rt Rt Reeeeevvvvvesionesionesionesionesion1. DEFINITION :
Complex numbers are definited as expressions of the form a + ib where a, b ∈ R & i = −1 . It isdenoted by z i.e. z = a + ib. ‘a’ is called as real part of z (Re z) and ‘b’ is called as imaginary part ofz (Im z).
EVERY COMPLEX NUMBER CAN BE REGARDED AS
Purely real Purely imaginary Imaginary if b = 0 if a = 0 if b ≠ 0
Note :(a) The set R of real numbers is a proper subset of the Complex Numbers. Hence the Complete Number
system is N ⊂ W ⊂ I ⊂ Q ⊂ R ⊂ C.(b) Zero is both purely real as well as purely imaginary but not imaginary.(c) i = −1 is called the imaginary unit. Also i² = − l ; i3 = −i ; i4 = 1 etc.(d) a b = a b only if atleast one of either a or b is non-negative.2. CONJUGATE COMPLEX :
If z = a + ib then its conjugate complex is obtained by changing the sign of its imaginary part &is denoted by z . i.e. z = a − ib.
Note that :(i) z + z = 2 Re(z) (ii) z − z = 2i Im(z) (iii) z z = a² + b² which is real(iv) If z lies in the 1st quadrant then z lies in the 4th quadrant and − z lies in the 2nd quadrant.3. ALGEBRAIC OPERATIONS :
The algebraic operations on complex numbers are similiar to those on real numbers treating i as apolynomial. Inequalities in complex numbers are not defined. There is no validity if we say that complexnumber is positive or negative.e.g. z > 0, 4 + 2i < 2 + 4 i are meaningless .However in real numbers if a2 + b2 = 0 then a = 0 = b but in complex numbers,z1
2 + z22 = 0 does not imply z1 = z2 = 0.
4. EQUALITY IN COMPLEX NUMBER :Two complex numbers z1 = a1 + ib1 & z2 = a2 + ib2 are equal if and only if their real & imaginaryparts coincide.
5. REPRESENTATION OF A COMPLEX NUMBER IN VARIOUS FORMS :(a) Cartesian Form (Geometric Representation) :
Every complex number z = x + i y can be represented by a point onthe cartesian plane known as complex plane (Argand diagram) by theordered pair (x, y).length OP is called modulus of the complex number denoted by z &θ is called the argument or amplitude .eg. z = x y2 2+ &
θ = tan−1 yx (angle made by OP with positive x−axis)
NOTE :(i) z is always non negative . Unlike real numbers z = z if zz if z
>− <
00
is not correct
(ii) Argument of a complex number is a many valued function . If θ is the argument of a complex numberthen 2 nπ + θ ; n ∈ I will also be the argument of that complex number. Any two arguments of acomplex number differ by 2nπ.
(iii) The unique value of θ such that – π < θ ≤ π is called the principal value of the argument.(iv) Unless otherwise stated, amp z implies principal value of the argument.(v) By specifying the modulus & argument a complex number is defined completely. For the complex number
0 + 0 i the argument is not defined and this is the only complex number which is given by its modulus.(vi) There exists a one-one correspondence between the points of the plane and the members of the set of
complex numbers.
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om(b) Trignometric / Polar Representation :
z = r (cos θ + i sin θ) where | z | = r ; arg z = θ ; z = r (cos θ − i sin θ)Note: cos θ + i sin θ is also written as CiS θ.
Also cos x = 2ee ixix −+
& sin x = 2ee ixix −−
are known as Euler's identities.(c) Exponential Representation :
z = reiθ ; | z | = r ; arg z = θ ; z = re− iθ
6. IMPORTANT PROPERTIES OF CONJUGATE / MODULI / AMPLITUDE :If z , z1 , z2 ∈ C then ;
(a) z + z = 2 Re (z) ; z − z = 2 i Im (z) ; )z( = z ; 21 zz + = 1z + 2z ;
21 zz − = 1z − 2z ; 21zz = 1z . 2z
2
1zz
= 2
1zz
; z2 ≠ 0
(b) | z | ≥ 0 ; | z | ≥ Re (z) ; | z | ≥ Im (z) ; | z | = | z | = | – z | ; z z = 2|z| ;
| z1 z2 | = | z1 | . | z2 | ; 2
1
zz
= |z||z|
2
1 , z2 ≠ 0 , | zn | = | z |n ;
| z1 + z2 |2 + | z1 – z2 |2 = 2 ][ 22
21 |z||z| +
z1− z2 ≤ z1 + z2 ≤ z1 + z2 [ TRIANGLE INEQUALITY ](c) (i) amp (z1 . z2) = amp z1 + amp z2 + 2 kπ. k ∈ I
(ii) amp zz
1
2
= amp z1 − amp z2 + 2 kπ ; k ∈ I
(iii) amp(zn) = n amp(z) + 2kπ .where proper value of k must be chosen so that RHS lies in (− π , π ].
(7) VECTORIAL REPRESENTATION OF A COMPLEX :Every complex number can be considered as if it is the position vector of that point. If the point P
represents the complex number z then, →
OP = z & →
OP = z.NOTE :
(i) If →
OP = z = r ei θ then →
OQ = z1 = r ei (θ + φ) = z . e iφ. If →
OP and →
OQ areof unequal magnitude then φ
ΛΛ= ieOPOQ
(ii) If A, B, C & D are four points representing the complex numbersz1, z2 , z3 & z4 then
AB CD if 12
34zzzz
−−
is purely real ; AB ⊥ CD if 12
34
zzzz
−−
is purely imaginary ](iii) If z1, z2, z3 are the vertices of an equilateral triangle where z0 is its circumcentre then
(a) z 12 + z 2
2 + z 32 − z1 z2 − z2 z3 − z3 z1 = 0 (b) z 1
2 + z 22 + z 3
2 = 3 z 02
8. DEMOIVRE’S THEOREM : Statement : cos n θ + i sin n θ is the value or one of the valuesof (cos θ + i sin θ)n ¥ n ∈ Q. The theorem is very useful in determining the roots of any complexquantity Note : Continued product of the roots of a complex quantity should be determined
using theory of equations.
9. CUBE ROOT OF UNITY : (i) The cube roots of unity are 1 , 2
3i1+− , 2
3i1−−.
(ii) If w is one of the imaginary cube roots of unity then 1 + w + w² = 0. In general1 + wr + w2r = 0 ; where r ∈ I but is not the multiple of 3.
(iii) In polar form the cube roots of unity are :
cos 0 + i sin 0 ; cos3
2π + i sin
32π
, cos3
4π + i sin
34π
(iv) The three cube roots of unity when plotted on the argand plane constitute the verties of an equilateral triangle.(v) The following factorisation should be remembered :
(a, b, c ∈ R & ω is the cube root of unity)
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oma3 − b3 = (a − b) (a − ωb) (a − ω²b) ; x2 + x + 1 = (x − ω) (x − ω2) ;a3 + b3 = (a + b) (a + ωb) (a + ω2b) ;a3 + b3 + c3 − 3abc = (a + b + c) (a + ωb + ω²c) (a + ω²b + ωc)
10. nth ROOTS OF UNITY :If 1 , α1 , α2 , α3 ..... αn − 1 are the n , nth root of unity then :(i) They are in G.P. with common ratio ei(2π/n) &(ii) 1p + α 1
p + α 2p + .... +α n
p−1 = 0 if p is not an integral multiple of n = n if p is an integral multiple of n
(iii) (1 − α1) (1 − α2) ...... (1 − αn − 1) = n &(1 + α1) (1 + α2) ....... (1 + αn − 1) = 0 if n is even and 1 if n is odd.
(iv) 1 . α1 . α2 . α3 ......... αn − 1 = 1 or −1 according as n is odd or even.11. THE SUM OF THE FOLLOWING SERIES SHOULD BE REMEMBERED :
(i) cos θ + cos 2 θ + cos 3 θ + ..... + cos n θ = ( )( )2sin
2nsinθθ
cos
+
21n
θ.
(ii) sin θ + sin 2 θ + sin 3 θ + ..... + sin n θ = ( )( )2sin
2nsinθθ
sin
+
21n
θ.Note : If θ = (2π/n) then the sum of the above series vanishes.
12. STRAIGHT LINES & CIRCLES IN TERMS OF COMPLEX NUMBERS :
(A) If z1 & z2 are two complex numbers then the complex number z = nm
mznz 21+
+ divides the joins of z1
& z2 in the ratio m : n.Note:(i) If a , b , c are three real numbers such that az1 + bz2 + cz3 = 0 ;
where a + b + c = 0 and a,b,c are not all simultaneously zero, then the complex numbers z1 , z2 & z3are collinear.(ii) If the vertices A, B, C of a ∆ represent the complex nos. z1, z2, z3 respectively, then :
(a) Centroid of the ∆ ABC = 3
zzz 321 ++ :
(b) Orthocentre of the ∆ ABC =( ) ( ) ( )
CseccBsecbAsecazCsecczBsecbzAseca 321
++++
OR CtanBtanAtan
CtanzBtanzAtanz 321
++++
(c) Incentre of the ∆ ABC = (az1 + bz2 + cz3) ÷ (a + b + c) .(d) Circumcentre of the ∆ ABC = :
(Z1 sin 2A + Z2 sin 2B + Z3 sin 2C) ÷ (sin 2A + sin 2B + sin 2C) .(B) amp(z) = θ is a ray emanating from the origin inclined at an angle θ to the x− axis.(C) z − a = z − b is the perpendicular bisector of the line joining a to b.(D) The equation of a line joining z1 & z2 is given by ;
z = z1 + t (z1 − z2) where t is a perameter.(E) z = z1 (1 + it) where t is a real parameter is a line through the point z1 & perpendicular to oz1.(F) The equation of a line passing through z1 & z2
can be expressed in the determinant form as
1zz1zz1zz
22
11 = 0. This is also the condition for three complex numbers to be collinear.
(G) Complex equation of a straight line through two given points z1 & z2 can be written as( ) ( ) ( )21212121 zzzzzzzzzz −+−−− = 0, which on manipulating takes the form as rzz +α+α = 0
where r is real and α is a non zero complex constant.(H) The equation of circle having centre z0 & radius ρ is :
z − z0 = ρ or z z − z0 z − 0z z +
0z z0 − ρ² = 0 which is of the form
rzzzz +α+α+ = 0 , r is real centre − α & radius r−αα .Circle will be real if 0r ≥−αα .
(I) The equation of the circle described on the line segment joining z1 & z2 as diameter is :
(i) arg1
2zzzz
−−
= ± 2π
or (z − z1) ( z − z 2)
+ (z − z2) ( z − z 1) = 0
(J) Condition for four given points z1 , z2 , z3 & z4 to be concyclic is, the number
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17 of
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14
24
23
13
zzzz
.zzzz
−−
−−
is real. Hence the equation of a circle through 3 non collinear points z1, z2 & z3 can be
taken as ( )( )( )( )231
132
zzzzzzzz
−−−−
is real ⇒ ( )( )( )( )231
132
zzzzzzzz
−−−−
= ( )( )( )( )231
132
zzzzzzzz
−−−−
13.(a) Reflection points for a straight line :Two given points P & Q are the reflection points for a given straight line if the given line is the rightbisector of the segment PQ. Note that the two points denoted by the complex numbers z1 & z2 will bethe reflection points for the straight line 0rzz =+α+α if and only if ; 0rzz 21 =+α+α , where r isreal and α is non zero complex constant.
(b) Inverse points w.r.t. a circle :Two points P & Q are said to be inverse w.r.t. a circle with centre 'O' and radius ρ, if :(i) the point O, P, Q are collinear and on the same side of O. (ii) OP . OQ = ρ2.Note that the two points z1 & z2 will be the inverse points w.r.t. the circle
0rzzzz =+α+α+ if and only if 0rzzzz 2121 =+α+α+ .14. PTOLEMY’S THEOREM : It states that the product of the lengths of the diagonals of a
convex quadrilateral inscribed in a circle is equal to the sum of the lengths of the two pairs ofits opposite sides. i.e. z1 − z3 z2 − z4 = z1 − z2 z3 − z4 + z1 − z4 z2 − z3 .
15. LOGARITHM OF A COMPLEX QUANTITY :
(i) Loge (α + i β) = 21
Loge (α² + β²) + i
αβ+π −1tann2 where n ∈ I.
(ii) ii represents a set of positive real numbers given by
π+π−
2n2
e , n ∈ I.VERY ELEMENTARY EXERCISE
Q.1 Simplify and express the result in the form of a + bi
(a) 2
i2i21
++ (b) −i (9 + 6 i) (2 − i)−1 (c)
23
1i2ii4
+− (d)
i52i23
i52i23
+−+
−+ (e) ( ) ( )
i2i2
i2i2 22
+−−
−+
Q.2 Given that x , y ∈ R, solve : (a) (x + 2y) + i (2x − 3y) = 5 − 4i (b) (x + iy) + (7 − 5i) = 9 + 4i(c) x² − y² − i (2x + y) = 2i (d) (2 + 3i) x² − (3 − 2i) y = 2x − 3y + 5i(e) 4x² + 3xy + (2xy − 3x²)i = 4y² − (x2/2) + (3xy − 2y²)i
Q.3 Find the square root of : (a) 9 + 40 i (b) −11 − 60 i (c) 50 iQ.4 (a) If f (x) = x4 + 9x3 + 35x2 − x + 4, find f ( – 5 + 4i)
(b) If g (x) = x4 − x3 + x2 + 3x − 5, find g(2 + 3i)Q.5 Among the complex numbers z satisfying the condition z i+ − =3 3 3 , find the number having the
least positive argument.Q.6 Solve the following equations over C and express the result in the form a + ib, a, b ∈ R.
(a) ix2 − 3x − 2i = 0 (b) 2 (1 + i) x2 − 4 (2 − i) x − 5 − 3 i = 0Q.7 Locate the points representing the complex number z on the Argand plane:
(a) z + 1 − 2i = 7 ; (b) z z− + +1 12 2 = 4 ; (c) zz
−+
33
= 3 ; (d) z − 3 = z − 6
Q.8 If a & b are real numbers between 0 & 1 such that the points z1 = a + i, z2 = 1 + bi & z3 = 0 form anequilateral triangle, then find the values of 'a' and 'b'.
Q.9 For what real values of x & y are the numbers − 3 + ix2 y & x2 + y + 4i conjugate complex?Q.10 Find the modulus, argument and the principal argument of the complex numbers.
(i) 6 (cos 310° − i sin 310°) (ii) −2 (cos 30° + i sin 30°) (iii) 24 1 2
++ +
ii i( )
Q.11 If (x + iy)1/3 = a + bi ; prove that 4 (a2 − b2) = xa
yb
+ .
Q.12(a) If a ibc id
++
= p + qi , prove that p2 + q2 = a bc d
2 2
2 2++
. (b) Let z1, z2, z3 be the complex numbers such that
z1 + z2 + z3 = z1z2 + z2z3 + z3z1 = 0. Prove that | z1 | = | z2 | = | z3 |.
Q.13 Let z be a complex number such that z ∈ c\R and 2
2
zz1zz1
+−++
∈ R, then prove that | z | =1.
Q.14 Prove the identity, ( ) ( )22
21
221
221 |z|1|z|1|zz||zz1| −−=−−−
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18 of
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Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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e at
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site:
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itjee
.com
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ww
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clas
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omQ.15 For any two complex numbers, prove that z z z z1 2
21 2
2+ + − = 2 [ ]z z12
22+ . Also give the
geometrical interpretation of this identity.Q.16 (a) Find all non−zero complex numbers Z satisfying Z = i Z².
(b) If the complex numbers z1, z2, .................zn lie on the unit circle |z| = 1 then show that|z1 + z2 + ..............+zn| = |z1
–1+ z2–1+................+zn
–1| .Q.17 Find the Cartesian equation of the locus of 'z' in the complex plane satisfying, | z – 4 | + | z + 4 | = 16.Q.18 If ω is an imaginary cube root of unity then prove that :
(a) (1 + ω − ω²)3 − (1− ω + ω²)3 = 0 (b) (1 − ω + ω²)5 + (1+ ω − ω²)5 = 32(c) If ω is the cube root of unity, Find the value of, (1 + 5ω2 + ω4) (1 + 5ω4 + ω2) (5ω3 + ω + ω2).
Q.19 If ω is a cube root of unity, prove that ; (i) (1 + ω − ω2)3 − (1 − ω + ω2)3
(ii) a b cc a b
+ ++ +
ω ωω ω
2
2 = ω2 (iii) (1 − ω) (1 − ω2) (1 − ω4) (1 − ω8) = 9
Q.20 If x = a + b ; y = aω + bω2 ; z = aω2 + bω, show that(i) xyz = a3 + b3 (ii) x2 + y2 + z2 = 6ab (iii) x3 + y3 + z3 = 3 (a3 + b3)
Q.21 If (w ≠ 1) is a cube root of unity then 11wii
1w1i1wwi112
22
−−+−−−−−
++ =
(A) 0 (B) 1 (C) i (D) wQ.22(a) (1 + w)7 = A + Bw where w is the imaginary cube root of a unity and A, B ∈ R, find the ordered pair
(A, B). (b) The value of the expression ;
1. (2 − w) (2 − w²) + 2. (3 − w) (3 − w²) + ............. + (n − 1) . (n − w) (n − w²), where w is animaginary cube root of unity is ________.
Q.23 If n ∈ N, prove that (1 + i)n + (1 − i)n = 2 2 1n + . cos nπ4
.
Q.24 Show that the sum k
n
=∑
1
2
sin cos22 1
22 1
π πkn
i kn+
−+
simplifies to a pure imaginary number.
Q.25 If x = cos θ + i sin θ & 1 + 1 2− a = na, prove that 1 + a cos θ = an2
(1 + nx) 1 +
nx
.Q.26 The number t is real and not an integral multiple of π/2. The complex number x1 and x2 are the roots of
the equation, tan2(t) · x2 + tan (t) · x + 1 = 0
Show that (x1)n + (x2)n =
π
3n2cos2 cotn(t).
EXEREXEREXEREXEREXERCISE-1CISE-1CISE-1CISE-1CISE-1Q.1 Simplify and express the result in the form of a + bi :
(a) −i (9 + 6 i) (2 − i)−1 (b) 23
1i2ii4
+− (c)
i52i23
i52i23
+−+
−+
(d) ( ) ( )i2
i2i2i2 22
+−−
−+ (e) ii −+
Q.2 Find the modulus , argument and the principal argument of the complex numbers.
(i) z = 1 + cos
π
910 + i sin
π
910
(ii) (tan1 – i)2
(iii) z = i125i125i125i125
−−+−++ (iv)
52sin
52cos1i
1iπ+
π−
−
Q.3 Given that x, y ∈ R, solve :
(a) (x + 2y) + i (2x − 3y) = 5 − 4i (b) 1i8i65
i23y
i21x
−+=
++
+(c) x² − y² − i (2x + y) = 2i (d) (2 + 3i) x² − (3 − 2i) y = 2x − 3y + 5i(e) 4x² + 3xy + (2xy − 3x²)i = 4y² − (x2/2) + (3xy − 2y²)i
Q.4(a) Let Z is complex satisfying the equation, z2 – (3 + i)z + m + 2i = 0, where m ∈ R.
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19 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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web
site:
ww
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tjeei
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.com
, w
ww
.teko
clas
ses.c
omSuppose the equation has a real root, then find the value of m.
(b) a, b, c are real numbers in the polynomial, P(Z) = 2Z4 + aZ3 + bZ2 + cZ + 3If two roots of the equation P(Z) = 0 are 2 and i, then find the value of 'a'.
Q.5(a) Find the real values of x & y for which z1 = 9y2 − 4 − 10 i x andz2 = 8y2 − 20 i are conjugate complex of each other.
(b) Find the value of x4 − x3 + x2 + 3x − 5 if x = 2 + 3i
Q.6 Solve the following for z : (a) z2 – (3 – 2 i)z = (5i – 5) (b) z + z = 2 + iQ.7(a) If i Z3 + Z2 − Z + i = 0, then show that | Z | = 1.
(b) Let z1 and z2 be two complex numbers such that 21
21
zz2z2z
−−
= 1 and | z2 | ≠ 1, find | z1 |.
(c) Let z1 = 10 + 6i & z2 = 4 + 6i. If z is any complex number such that the argument of, 2
1zzzz
−−
is 4π, then
prove that z − 7 − 9i = 3 2 .Q.8 Show that the product,
++
++
++
++
n2 222
2i11......
2i11
2i11
2i11 is equal to 1 1
22−
n (1+ i) where n ≥ 2 .
Q.9 Let a & b be complex numbers (which may be real) and let,Z = z3 + (a + b + 3i) z2 + (ab + 3 ia + 2 ib − 2) z + 2 abi − 2a.(i) Show that Z is divisible by, z + b + i. (ii) Find all complex numbers z for which Z = 0.(iii) Find all purely imaginary numbers a & b when z = 1 + i and Z is a real number.
Q.10 Interpret the following locii in z ∈ C.
(a) 1 < z − 2i < 3 (b) Re 42zii2z ≤
+
+ (z ≠ 2i)
(c) Arg (z + i) − Arg (z − i) = π/2 (d) Arg (z − a) = π/3 where a = 3 + 4i.Q.11 Prove that the complex numbers z1 and z2 and the origin form an isosceles triangle with vertical angle
2π/3 if 0zzzz 2122
21 =++ .
Q.12 P is a point on the Aragand diagram. On the circle with OP as diameter two points Q & R are taken suchthat ∠ POQ = ∠ QOR = θ. If ‘O’ is the origin & P, Q & R are represented by the complex numbersZ1 , Z2 & Z3 respectively, show that : Z2
2 . cos 2 θ = Z1 . Z3 cos² θ.
Q.13 Let z1, z2, z3 are three pair wise distinct complex numbers and t1, t2, t3 are non-negative real numberssuch that t1 + t2 + t3 = 1. Prove that the complex number z = t1z1 + t2z2 + t3z3 lies inside a triangle withvertices z1, z2, z3 or on its boundry.
Q.14 If a CiS α , b CiS β , c CiS γ represent three distinct collinear points in an Argand's plane, then provethe following :(i) Σ ab sin (α − β) = 0.(ii) (a CiS α) )cos(bc2cb 22 γ−β−+ ± (b CiS β) )cos(ac2ca 22 γ−α−+
∓ (c CiS γ) )cos(ab2ba 22 β−α−+ = 0.Q.15 Find all real values of the parameter a for which the equation
(a − 1)z4 − 4z2 + a + 2 = 0 has only pure imaginary roots.
Q.16 Let A ≡ z1 ; B ≡ z2; C ≡ z3 are three complex numbers denoting the vertices of an acute angled triangle.If the origin ‘O’ is the orthocentre of the triangle, then prove that
z1 z2 + z1 z2 = z2 z3 + z2 z3 = z3 z1 + z3 z1hence show that the ∆ ABC is a right angled triangle ⇔ z1 z2 + z1 z2 = z2 z3 + z2 z3 = z3 z1 + z3 z1 = 0
Q.17 If the complex number P(w) lies on the standard unit circle in an Argand's plane andz = (aw+ b)(w – c)–1 then, find the locus of z and interpret it. Given a, b, c are real.
Q.18(a) Without expanding the determinant at any stage , find RK∈ such that
i8iKi4ii16i8
i34i8i4
+−+−
++ has purely imaginary value.
(b) If A, B and C are the angles of a triangle
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20 of
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D = iC2iAiB
iAiB2iC
iBiCiA2
eeeeeeeee
−
−
−
where i = −1 then find the value of D.
Q.19 If w is an imaginary cube root of unity then prove that :(a) (1 − w + w2) (1 − w2 + w4) (1 − w4 + w8) ..... to 2n factors = 22n .(b) If w is a complex cube root of unity, find the value of
(1 + w) (1 + w2) (1 + w4) (1 + w8) ..... to n factors .
Q.20 Prove that n
cosisin1cosisin1
θ−θ+θ+θ+ = cos
θ−π n
2n
+ i sin
θ−π n
2n
. Hence deduce that5
5cosi
5sin1
π+π+ + i
5
5cosi
5sin1
π−π+ = 0
Q.21 If cos (α − β) + cos (β − γ) + cos (γ − α) = − 3/2 then prove that :(a) Σ cos 2α = 0 = Σ sin 2α (b) Σ sin (α + β) = 0 = Σ cos (α + β) (c) Σ sin2 α = Σ cos2 α = 3/2(d) Σ sin 3α = 3 sin (α + β + γ) (e) Σ cos 3α = 3 cos (α + β + γ)(f) cos3 (θ + α) + cos3 (θ + β) + cos3 (θ + γ) = 3 cos (θ + α) . cos (θ + β) . cos (θ + γ) where θ ∈ R.Q.22 Resolve Z5 + 1 into linear & quadratic factors with real coefficients. Deduce that : 4·sin π
10·cos π
5= 1.
Q.23 If x = 1+ i 3 ; y = 1 − i 3 & z = 2 , then prove that xp + yp = zp for every prime p > 3.Q.24 If the expression z5 – 32 can be factorised into linear and quadratic factors over real coefficients as
(z5 – 32) = (z – 2)(z2 – pz + 4)(z2 – qz + 4) then find the value of (p2 + 2p).Q.25(a) Let z = x + iy be a complex number, where x and y are real numbers. Let A and B be the sets defined by
A = {z | | z | ≤ 2} and B = {z | (1 – i)z + (1 + i) z ≥ 4}. Find the area of the region A ∩ B.
(b) For all real numbers x, let the mapping f (x) = i−x
1, where i = 1− . If there exist real number
a, b, c and d for which f (a), f (b), f (c) and f (d) form a square on the complex plane. Find the area ofthe square.
EXEREXEREXEREXEREXERCISE-2CISE-2CISE-2CISE-2CISE-2Q.1 If
p q rq r pr p q
= 0 ; where p , q , r are the moduli of non−zero complex numbers u, v, w respectively,
prove that, arg wv
= arg w uv u
−−
2
.
Q.2 The equation x3 = 9 + 46i where i = 1− has a solution of the form a + bi where a and b are integers.Find the value of (a3 + b3).
Q.3 Show that the locus formed by z in the equation z3 + iz = 1 never crosses the co-ordinate axes in the
Argand’s plane. Further show that |z| = −
+Im( )
Re( ) Im( )z
z z2 1Q.4 If ω is the fifth root of 2 and x = ω + ω2, prove that x5 = 10x2 + 10x + 6.Q.5 Prove that , with regard to the quadratic equation z2 + (p + ip′) z + q + iq′ = 0
where p , p′, q , q′ are all real.(i) if the equation has one real root then q ′2 − pp ′ q ′ + qp ′2 = 0 .(ii) if the equation has two equal roots then p2 − p′2 = 4q & pp ′ = 2q ′.
State whether these equal roots are real or complex.Q.6 If the equation (z + 1)7 + z7 = 0 has roots z1, z2, .... z7, find the value of
(a) ∑=
7
1rr )ZRe( and (b) ∑
=
7
1rr )ZIm(
Q.7 Find the roots of the equation Zn = (Z + 1)n and show that the points which represent them are collinearon the complex plane. Hence show that these roots are also the roots of the equation
22
Zn
msin2
π + Z
nmsin2
2
π + 1 = 0.
Q.8 Dividing f(z) by z − i, we get the remainder i and dividing it by z + i, we get the remainder
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om1 + i. Find the remainder upon the division of f(z) by z² + 1.
Q.9 Let z1 & z2 be any two arbitrary complex numbers then prove that :
z1 + z2 ≥ ( )|z|
z|z|
z|z||z|
21
2
2
1
121 ++ .
Q.10 If Zr, r = 1, 2, 3, ......... 2m, m ε N are the roots of the equation
Z2m + Z2m-1 + Z2m-2 + ............. + Z + 1 = 0 then prove that ∑−=r
m
rZ1
2 11 = − m
Q.11 If (1 + x)n = C0 + C1x + C2x² + .... + Cn xn (n ∈ N), prove that :
(a) C0 + C4
+ C8 + .... = 1
22 2
41 2n n n− +
/ cos π (b) C1 + C5
+ C9 + .... = 12
2 24
1 2n n n− +
/ sin π
(c) C2 + C6
+ C10 + ..... = 1
22 2
41 2n n n− −
/ cos π (d) C3 + C7
+ C11 + .... = 1
22 2
41 2n n n− −
/ sin π
(e) C0 + C3 + C6 + C9 + ........ = 13
2 23
n n+
cos π
Q.12 Let z1 , z2 , z3 , z4 be the vertices A , B , C , D respectively of a square on the Argand diagramtaken in anticlockwise direction then prove that :(i) 2z2 = (1 + i) z1 + (1− i)z3 & (ii) 2z4 = (1− i) z1 + (1 + i) z3
Q.13 Show that all the roots of the equation 11
11
+−
= +
−i xi x
i ai a
n
a ∈ R are real and distinct.
Q.14 Prove that:
(a) cos x + nC1 cos 2x + nC2 cos 3x + ..... + nCn cos (n + 1) x = 2n . cosn x2
. cos n +
22
x
(b) sin x + nC1 sin 2x + nC2 sin 3x + ..... + nCn sin (n + 1) x = 2n . cosn x2
. sin n +
22
x
(c) cos 22 1
πn +
+ cos 4
2 1π
n +
+ cos 6
2 1π
n +
+ ..... + cos 2
2 1n
nπ+
= − 1
2 When n ∈ N.
Q.15 Show that all roots of the equation a0zn + a1z
n – 1 + ...... + an – 1z + an = n,
where | ai | ≤ 1, i = 0, 1, 2, .... , n lie outside the circle with centre at the origin and radius n1n −
.Q.16 The points A, B, C depict the complex numbers z1 , z2 , z3 respectively on a complex plane & the angle
B & C of the triangle ABC are each equal to )(21 α−π . Show that
(z2 − z3)² = 4 (z3 − z1) (z1 − z2) sin2 α2
.
Q.17 Show that the equation Ax a
Ax a
Ax a
n
n
12
1
22
2
2
−+
−+ +
−...... = k has no imaginary root, given that:
a1 , a2 , a3 .... an & A1, A2, A3 ..... An, k are all real numbers.
Q.18 Let a, b, c be distinct complex numbers such that b1a− = c1
b− = a1
c− = k. Find the value of k.
Q.19 Let α, β be fixed complex numbers and z is a variable complex number such that,z − α 2 + z − β 2 = k.
Find out the limits for 'k' such that the locus of z is a circle. Find also the centre and radius of the circle.Q.20 C is the complex number. f : C → R is defined by f (z) = | z3 – z + 2|. What is the maximum value of f on
the unit circle | z | = 1?
Q.21 Let f (x) = )xi2(coslogx3cos if x ≠ 0 and f (0) = K (where i = 1− ) is continuous at x = 0 then find
the value of K. Use of L Hospital’s rule or series expansion not allowed.
Q.22 If z1 , z2 are the roots of the equation az2 + bz + c = 0, with a, b, c > 0 ; 2b2 > 4ac > b2 ;z1 ∈ third quadrant ; z2 ∈ second quadrant in the argand's plane then, show that
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22 of
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arg
2
1
zz
= 2cos–1
2/12
ac4b
Q.23 Find the set of points on the argand plane for which the real part of the complex number(1 + i) z2 is positive where z = x + iy , x, y ∈ R and i = −1 .
Q.24 If a and b are positive integer such that N = (a + ib)3 – 107i is a positive integer. Find N.Q.25 If the biquadratic x4 + ax3 + bx2 + cx + d = 0 (a, b, c, d ∈ R) has 4 non real roots, two with sum
3 + 4i and the other two with product 13 + i. Find the value of 'b'.
EXEREXEREXEREXEREXERCISE-3CISE-3CISE-3CISE-3CISE-3Q.1 Evaluate: ( ) sin cos3 2 2
112111
10
1
32p q i q
q
p
p+ −
==∑∑ π π . [REE '97, 6]
Q.2(a) Let z1 and z2 be roots of the equation z2 + pz + q = 0 , where the co−efficients p and q may becomplex numbers. Let A and B represent z1 and z2 in the complex plane. If ∠ AOB = α ≠ 0 and
OA = OB, where O is the origin . Prove that p2 = 4 q cos2 α2
. [JEE '97 , 5]
(b) Prove that k
n
=
−
∑1
1
(n − k) cos2 k
nπ
= − n2
where n ≥ 3 is an integer . [JEE '97, 5]
Q.3(a) If ω is an imaginary cube root of unity, then (1 + ω − ω2)7 equals(A) 128ω (B) − 128ω (C) 128ω2 (D) − 128ω2
(b) The value of the sum ( )i in n
n+ +
=∑ 1
1
13 , where i = −1 , equals
(A) i (B) i − 1 (C) − i (D) 0 [JEE' 98, 2 + 2 ]Q.4 Find all the roots of the equation (3z − 1)4 + (z − 2)4 = 0 in the simplified form of a + ib.
[REE ’98, 6 ]
Q.5(a) If i = −1 , then 4 + 5 − +
12
32
334i
+ 3 − +
12
32
365i
is equal to :
(A) 1 − i 3 (B) − 1 + i 3 (C) i 3 (D) − i 3
(b) For complex numbers z & ω, prove that, z 2 ω − ω 2 z = z − ω if and only if, z = ω or z ω = 1 [JEE '99, 2 + 10 (out of 200)]
Q.6 If α = ei2
7π
and f(x) = A0 + k =∑
1
20
Ak xk, then find the value of,
f(x) + f(α x) + ...... + f(α6x) independent of α . [REE '99, 6]
Q.7(a) If z1 , z2 , z3 are complex numbers such that z1 = z2 = z3 = 1 1 1
1 2 3z z z+ +
= 1, then
z1 + z2 + z3 is :(A) equal to 1 (B) less than 1 (C) greater than 3 (D) equal to 3
(b) If arg (z) < 0 , then arg (− z) − arg (z) =(A) π (B) − π (C) −
π2
(D) π2
[ JEE 2000 (Screening) 1 + 1 out of 35 ]
Q.8 Given , z = cos 2
2 1π
n + + i sin 2
2 1π
n + , 'n' a positive integer, find the equation whose roots are,α = z + z3 + ...... + z2n − 1 & β = z2 + z4 + ...... + z2n .
[ REE 2000 (Mains) 3 out of 100 ]
Q.9(a) The complex numbers z1, z2 and z3 satisfying z zz z
i1 3
2 3
1 32
−−
= − are the vertices of a triangle which is
(A) of area zero (B) right-angled isosceles(C) equilateral (D) obtuse – angled isosceles
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e at
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site:
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itjee
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ww
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clas
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om (b) Let z1 and z2 be nth roots of unity which subtend a right angle at the origin. Then n must be of the form
(A) 4k + 1 (B) 4k + 2 (C) 4k + 3 (D) 4k[ JEE 2001 (Scr) 1 + 1 out of 35 ]
Q.10 Find all those roots of the equation z12 – 56z6 – 512 = 0 whose imaginary part is positive.[ REE 2000, 3 out of 100 ]
Q.11(a) Let ω = − +12
32
i . Then the value of the determinant 1 1 11 11
2 2
2 4
− − ω ωω ω
is
(A) 3ω (B) 3ω (ω – 1) (C) 3ω2 (D) 3ω(1 – ω) (b) For all complex numbers z1, z2 satisfying |z1| = 12 and |z2 – 3 – 4i| = 5, the minimum value of
|z1 – z2| is(A) 0 (B) 2 (C) 7 (D) 17
[JEE 2002 (Scr) 3+3] (c) Let a complex number α , α ≠ 1, be a root of the equation
zp+q – zp – zq + 1 = 0 where p, q are distinct primes.Show that either 1 + α + α2 + ...... + αp–1 = 0 or 1 + α + α2 + ...... + αq–1 = 0 , but not both together.
[JEE 2002, (5) ]
Q.12(a) If z1 and z2 are two complex numbers such that | z1 | < 1 < | z2 | then prove that 1zzzz1
21
21 <−
−.
(b) Prove that there exists no complex number z such that | z | < 31
and ∑=
n
1r
rr za = 1 where | ar | < 2.
[JEE-03, 2 + 2 out of 60]Q.13(a) ω is an imaginary cube root of unity. If (1 + ω2)m = (1 + ω4)m , then least positive integral value of m is
(A) 6 (B) 5 (C) 4 (D) 3[JEE 2004 (Scr)]
(b) Find centre and radius of the circle determined by all complex numbers z = x + i y satisfying k)z()z( =
β−α− ,
where 2121 i,i β+β=βα+α=α are fixed complex and k ≠ 1. [JEE 2004, 2 out of 60 ]
Q.14(a) The locus of z which lies in shaded region is best represented by(A) z : |z + 1| > 2, |arg(z + 1)| < π/4(B) z : |z - 1| > 2, |arg(z – 1)| < π/4(C) z : |z + 1| < 2, |arg(z + 1)| < π/2(D) z : |z - 1| < 2, |arg(z - 1)| < π/2
(b) If a, b, c are integers not all equal and w is a cube root of unity (w ≠ 1), then the minimum value of|a + bw + cw2| is
(A) 0 (B) 1 (C) 23
(D) 21
[JEE 2005 (Scr), 3 + 3] (c) If one of the vertices of the square circumscribing the circle |z – 1| = 2 is i32 + . Find the othervertices of square. [JEE 2005 (Mains), 4]
Q.15 If w = α + iβ where β ≠ 0 and z ≠ 1, satisfies the condition that z1zww
−−
is purely real, then the set ofvalues of z is(A) {z : | z | = 1} (B) {z : z = z ) (C) {z : z ≠ 1} (D) {z : | z | = 1, z ≠ 1}
[JEE 2006, 3]ANSWER KEY
VERY ELEMENTARY EXERCISEQ.1 (a)
2524
257 + i; (b)
512
521 − i; (c) 3 + 4i; (d)
298− + 0i; (e)
522 i
Q.2 (a) x =1, y = 2; (b) (2, 9); (c) (−2 , 2) or − −
23
23
, ; (d) (1 ,1) 0 52
,
(e) x = K, y = 32K , K∈ R
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24 of
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omQ.3 (a) ± (5 + 4i) ; (b) ± (5 − 6i) (c) ± 5(1 + i) Q.4 (a) −160 ; (b) − (77 +108 i)
Q.5 – 32
3 32
+ i Q.6 (a) − i , − 2i (b) 3 5
2− i
or − 12+ i
Q.7 (a) on a circle of radius 7 with centre (−1, 2) ; (b) on a unit circle with centre at origin(c) on a circle with centre (−15/4, 0) & radius 9/4 ; (d) a straight line
Q.8 a = b = 2 − 3 ; Q.9 x = 1, y = − 4 or x = − 1, y = − 4Q.10 (i) Modulus = 6 , Arg = 2 k π + 5
18π (K ∈ I) , Principal Arg = 5
18π (K ∈ I)
(ii) Modulus = 2 , Arg = 2 k π + 76π , Principal Arg = − 5
6π
(iii) Modulus = 56
, Arg = 2 k π − tan−1 2 (K ∈ I) , Principal Arg = − tan−12
Q.16 (a) 32 2
− i , − −32 2
i , i ; Q.17 148y
64x 22
=+ ; Q.18 (c) 64 ; Q.21 A
Q.22 (a) (1, 1) ; (b) ( )n nn
+
−1
2
2
EXEREXEREXEREXEREXERCISE-1CISE-1CISE-1CISE-1CISE-1Q.1 (a) 21
5125
− i (b) 3 + 4 i (c) − 829
+ 0 i (d) 225
i (e) + i02 + or i20±
Q.2 (i) Principal Arg z = − 49π
; z = 2 cos49π
; Arg z = 2 k π − 49π
k ∈ I (ii) Modulus = sec21 , Arg = 2 n π + (2 – π ) , Principal Arg = (2 – π )
(iii) Principal value of Agr z = − π2 & z =
32 ; Principal value of Arg z =
π2 & z =
23
(iv) Modulus = 5
eccos2
1 π , Arg z = 20
11n2 π+π , Principal Arg = 20
11π
Q.3(a) x = 1, y = 2; (b) x = 1 & y = 2 ; (c) (−2 , 2) or − −
23
23
, ; (d) (1 ,1) 0 52
,
; (e) x =K, y = 32K K∈ R
Q.4 (a) 2, (b) – 11/2 Q.5 (a) [(− 2, 2) ; (− 2, − 2)] (b) − (77 +108 i)Q.6 (a) z = (2 + i) or (1 – 3i); (b) z = 3 4
4+ i
Q.7 (b) 2
Q.9 (ii) z = − (b + i) ; − 2 i , − a (iii)
+− ti,
5t3ti2
where t ∈ R −
−
35
Q.10 (a) The region between the co encentric circles with centre at (0 , 2) & radii 1 & 3 units
(b) region outside or on the circle with centre 21
+ 2i and radius 21
.(c) semi circle (in the 1st & 4th quadrant) x² + y² = 1 (d) a ray emanating from the point (3 + 4i) directed away from the origin & having equation 3 4 3 3 0x y− + − =
Q.15 [−3 , −2] Q.17 (1 – c2) | z |2 – 2(a + bc) (Re z) + a2 – b2 = 0Q.18 (a) K = 3 , (b) – 4 Q.19 (b) one if n is even ; − w² if n is oddQ.22 (Z + 1) (Z² − 2Z cos 36° + 1) (Z² − 2Z cos 108° + 1) Q.24 4Q.25 (a) π – 2 ; (b) 1/2
EXEREXEREXEREXEREXERCISE-2CISE-2CISE-2CISE-2CISE-2Q.2 35 Q.6 (a) –
27 , (b) zero Q.8 i z i
212
+ + Q.18 – ω or – ω2
Q.19 k > 12
2α β− Q.20 | f (z) | is maximum when z = ω, where ω is the cube root unity and | f (z) | = 13
Q.21 K = – 94
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25 of
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Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
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omQ.23 required set is constituted by the angles without their boundaries, whose sides are the straight lines
y = )12( − x and y + )12( + x = 0 containing the x − axisQ.24 198 Q.25 51
EXEREXEREXEREXEREXERCISE-3CISE-3CISE-3CISE-3CISE-3Q.1 48(1 − i) Q.3 (a) D (b) B
Q.4 Z = ( ) ( )29 20 2 15 25 2
82+ + ± +i
, ( ) ( )29 20 2 15 25 2
82− + ± −i
Q.5 (a) C
Q.6 7 A0 + 7 A7 x7 + 7 A14 x
14 Q.7 (a) A (b) A Q.8 z2 + z + sinsin
2
2n θθ = 0, where θ = 2
2 1π
n +
Q.9 (a) C, (b) D Q.10 +1 + i 3 , ( )± +3
2
i, 2 i Q.11 (a) B ; (b) B
Q.13 (a) D ; (b) Centre ≡ 1k
k2
2
−α−β
, Radius = ( )( )1k.||||.k|k|)1k(
1 2222222 −α−β−β−α
−Q.14 (a) A, (b) B, (c) z2 = – 3 i ; z3 = ( ) i31 +− ; z4 = ( ) i31 −+ Q.15 D
EXEREXEREXEREXEREXERCISE-4CISE-4CISE-4CISE-4CISE-4Part : (A) Only one correct option
1. If |z| = 1 and ω = 1z1z
+−
(where z ≠ –1), the Re(ω) is [IIT – 2003, 3]
(A) 0 (B) 2|1z|1+
− (C) 2|1z|1.
1zz
++ (D) 2|1z|2
+2. The locus of z which lies in shaded region (excluding the boundaries) is best represented by
[IIT – 2005, 3]
(A) z : |z + 1| > 2 and |arg (z + 1)| < π/4 (B) z : |z – 1| > 2 and |arg (z – 1)| < π/4(C) z : |z + 1| < 2 and |arg (z + 1)| < π/2 (D) z : |z – 1| < 2 and |arg (z + 1)| < π/2
3. If w = α, + iβ, where β ≠ 0 and z ≠ 1, satisfies the condition that
−−
z1zww
is purely real, then the set ofvalues of z is [IIT – 2006, (3, –1)](A) {z : |z| = 1} (B) {z : z = z } (C) {z : z ≠ 1} (D) {z : |z| = 1, z ≠1}
4. If ( 3 + i)100 = 299 (a + ib), then b is equal to
(A) 3 (B) 2 (C) 1 (D) none of these
5. If Re
+−
6zi8z
= 0, then z lies on the curve(A) x2 + y2 + 6x – 8y = 0 (B) 4x – 3y + 24 = 0 (C) 4ab (D) none of these
6. If n1, n2 are positive integers then : 1n)i1( + + 1n3 )i1( + + 2n5 )i1( − + 2n7 )i1( − is a real number if and only if(A) n1 = n2 + 1 (B) n1 + 1 = n2(C) n1 = n2 (D) n1, n2 are any two positive integers
7. The three vertices of a triangle are represented by the complex numbers, 0, z1 and z2. If the triangle isequilateral, then(A) z1
2 – z22 = z1z2 (B) z2
2 – z12 = z1 z2 (C) z1
2 + z22 = z1z2 (D) z1
2 + z22 + z1z2 = 0
8. If x2 – x + 1 = 0 then the value of 25
1nn
n
x1x∑
=
+ is
(A) 8 (B) 10 (C) 12 (D) none of these
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26 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
FREE
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nloa
d St
udy
Pack
age
Vie
ws o
f stu
dent
s ava
ilabl
e at
web
site:
ww
w.ii
tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
om9. If α is nonreal and α = 5 1 then the value of |1| 122
2−− α−α+α+α+ is equal to
(A) 4 (B) 2 (C) 1 (D) none of these
10. If z = x + iy and z1/3 = a − ib then ( )xa
yb
k a b− = −2 2 where k =(A) 1 (B) 2 (C) 3 (D) 4
11. − +
+ − −
+ − +
+ − −
1 32
1 32
1 32
1 32
6 6 5 5i i i i is equal to :
(A) 1 (B) − 1 (C) 2 (D) none12. Expressed in the form r (cos θ + i sin θ), − 2 + 2i becomes :
(A) 2 24 4
cos sin−
+ −
π πi (B) 2 2 34
34
cos sinπ π
+
i
(C) 2 2 34
34
cos sin−
+ −
π πi (D) 24 4
cos sin−
+ −
π πi
13. The number of solutions of the equation in z, z z - (3 + i) z - (3 - i) z - 6 = 0 is :(A) 0 (B) 1 (C) 2 (D) infinite
14. If |z| = max {|z – 1|, |z + 1|} then
(A) |z + z | = 21
(B) z + z = 1 (C) |z + z | = 1 (D) none of these15. If P, P′ represent the complex number z1 and its additive inverse respectively then the complex equation of
the circle with PP′ as a diameter is
(A) 1z
z =
zz1 (B) z z + z1 1z = 0 (C) z 1z + z z1 = 0 (D) none of these
16. The points z1 = 3 + 3 i and z2 = 2 3 + 6 i are given on a complex plane. The complex number lyingon the bisector of the angle formed by the vectors z1 and z2 is :
(A) z =2
232
)323( +++ i (B) z = 5 + 5 i(C) z = − 1 − i (D) none
17. The expression 11
11
+−
− +
−ii
i ni n
ntantan
tantan
αα
αα
when simplified reduces to :
(A) zero (B) 2 sin n α (C) 2 cos n α (D) none18. All roots of the equation, (1 + z)6 + z6 = 0 :
(A) lie on a unit circle with centre at the origin (B)lie on a unit circle with centre at (− 1, 0)(C) lie on the vertices of a regular polygon with centre at the origin (D) are collinear
19. Points z1 & z2 are adjacent vertices of a regular octagon. The vertex z3 adjacent to z2 (z3 ≠ z1) isrepresented by :
(A) z2 +12
(1 ± i) (z1 + z2) (B) z2 +12
(1 ± i) (z1 − z2)
(C) z2 +12
(1 ± i) (z2 − z1) (D) none of these
20. If z = x + i y then the equation of a straight line Ax + By + C = 0 where A, B, C ∈ R, can be written onthe complex plane in the form a z a z C+ + 2 = 0 where 'a' is equal to :
(A) ( )A iB+
2(B)
A iB−2 (C) A + i B (D) none
21. The points of intersection of the two curves z − 3 = 2 and z = 2 in an argand plane are:
(A)12 ( )7 3± i (B)
12 ( )3 7± i (C)
32 ± i
72
(D)72 ± i
32
22. The equation of the radical axis of the two circles represented by the equations, z − 2 = 3 and z − 2 − 3 i = 4 on the complex plane is :(A) 3iz – 3i z – 2 = 0 (B) 3iz – 3i z + 2 = 0 (C) iz – i z + 1 = 0 (D) 2iz – 2i z + 3 = 0
23. Ifr
1p=Π eipθ = 1 where Π denotes the continued product, then the most general value of θ is :
(A) 2
1n
r rπ
( )−(B)
21
nr r
π( )+
(C) 4
1n
r rπ
( )−(D)
41
nr r
π( )+
24. The set of values of a ∈ R for which x2 + i(a – 1) x + 5 = 0 will have a pair of conjugate imaginary roots is(A) R (B) {1} (C) |a| a2 – 2a + 21 > 0} (D) none of these
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27 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
FREE
Dow
nloa
d St
udy
Pack
age
Vie
ws o
f stu
dent
s ava
ilabl
e at
web
site:
ww
w.ii
tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
om25. If |z1 – 1| < 1, |z2 – 2| < 2, |z3 – 3| < 3 then |z1 + z2 + z3|(A) is less than 6 (B) is more than 3
(C) is less than 12 (D) lies between 6 and 1226. If z1, z2, z3, ........., zn lie on the circle |z| = 2, then the value of
E = |z1 + z2 + ..... + zn| – 4 n21 z1.......
z1
z1 +++ is
(A) 0 (B) n (C) –n (D) none of thesePart : (B) May have more than one options correct27. If z1 lies on |z| = 1 and z2 lies on |z| = 2, then
(A) 3 ≤ |z1 – 2z2| ≤ 5 (B) 1 ≤ |z1 + z2| ≤ 3(C) |z1 – 3z2| ≥ 5 (D) |z1 – z2| ≥ 1
28. If z1, z2, z3, z4 are root of the equation a0z4 + z1z3 + z2z2 + z3z + z4 = 0, where a0, a1, a2, a3 and a4 are real,then(A) 1z , 2z , 3z , 4z are also roots of the equation (B) z1 is equal to at least one of 1z , 2z , 3z , 4z(C) – 1z ,– 2z , – 3z , – 4z are also roots of the equation (D) none of these
29. If a3 + b3 + 6 abc = 8 c3 & ω is a cube root of unity then :(A) a, c, b are in A.P. (B) a, c, b are in H.P.(C) a + bω − 2 cω2 = 0 (D) a + bω2 − 2 cω = 0
30. The points z1, z2, z3 on the complex plane are the vertices of an equilateral triangle if and only if :(A) Σ (z1 − z2) (z2 − z3) = 0 (B) z1
2 + z22 + z3
2 = 2 (z1 z2 + z2 z3 + z3 z1)(C) z1
2 + z22 + z3
2 = z1 z2 + z2 z3 + z3 z1 (D) 2 (z12 + z2
2 + z32) = z1 z2 + z2 z3 + z3 z1
31. If |z1 + z2| = |z1 – z2| then
(A) |amp z1 – amp z2| = 2π
(B) | amp z1 – amp2| = π
(C) 2
1
zz
is purely real (D) 2
1
zz
is purely imaginary
EXEREXEREXEREXEREXERCISE-5CISE-5CISE-5CISE-5CISE-51. Given that x, y ∈ R, solve : 4x² + 3xy + (2xy − 3x²)i = 4y² − (x2/2) + (3xy − 2y²)i
2. If α & β are any two complex numbers, prove that :
α α β α α β α β α β− − + + − = + + −2 2 2 2 .3. If α, β are the numbers between 0 and 1, such that the points z1 = α + i, z2 = 1 + βi and z3 = 0 form an
equilateral triangle, then find α and β.4. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the points D
and M represent the complex numbers 1 + i and 2 - i respectively, then find the complex number correspondingto A.
5. Show that the sum of the pth powers of nth roots of unity :(a) is zero, when p is not a multiple of n. (b) is equal to n, when p is a multiple of n.
6. If (1 + x)n = p0 + p1 x + p2 x2 + p3 x3 +......., then prove that :
(a) p0 − p2 + p4 −....... = 2n/2 cos n π4
(b) p1 − p3 + p5 −....... = 2n/2 sin n π4
7. Prove that, loge1
1 −
ei θ = loge
12 2
cosec θ
+ i
π θ2 2
−
8. If iii ....... ∞
= A + i B, principal values only being considered, prove that
(a) tan12
πA = BA
(b) A2 + B2 = e − π B
9. Prove that the roots of the equation, (x - 1)n = xn are12
1 +
i rr
cot π, where
r = 0, 1, 2,....... (n − 1) & n ∈ N.10. If cos (α − β) + cos (β − γ) + cos (γ − α) = − 3/2 then prove that :
(a) Σ cos 2α = 0 = Σ sin 2α (b) Σ sin (α + β) = 0 = Σ cos (α + β)(c) Σ sin 3α = 3 sin (α + β + γ) (d) Σ cos 3 α = 3 cos (α + β + γ)(e) Σ sin2 α = Σ cos2 α = 3/2(f) cos3 (θ + α) + cos3 (θ + β) + cos3 (θ + γ) = 3 cos (θ + α). cos (θ + β). cos (θ + γ)
where θ ∈ R.
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28 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
FREE
Dow
nloa
d St
udy
Pack
age
Vie
ws o
f stu
dent
s ava
ilabl
e at
web
site:
ww
w.ii
tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
om11. If α , β, γ are roots of x3 − 3 x2 + 3 x + 7 = 0 (and ω is imaginary cube root of unity), then find the value
ofαβ
−−
11
+βγ
−−
11
+γα
−−
11
.
12. Given that, |z − 1| = 1, where ' z ' is a point on the argand plane. Show that z
z− 2
= i tan (arg z).
13. P is a point on the Argand diagram. On the circle with OP as diameter two points Q & R are taken suchthat ∠ POQ = ∠ QOR = θ. If ‘O’ is the origin & P, Q & R are represented by the complex numbersZ1, Z2 & Z3 respectively, show that : Z2
2. cos 2 θ = Z1. Z3 cos² θ.
14. Find an expression for tan 7θ in terms of tan θ, using complex numbers. By consideringtan 7θ = 0, show that x = tan2 (3 π/7) satisfies the cubic equation x3 − 21x2 + 35x − 7 = 0.
15. If (1 + x)n = C0 + C1x + C2x² +.... + Cn xn (n ∈ N), prove that : C2 + C6
+ C10 +..... = 1
22 2
41 2n n n− −
/ cos π
16. Prove that : cos 22 1
πn +
+ cos 4
2 1π
n +
+ cos 6
2 1π
n +
+..... + cos
22 1
nn
π+
= − 1
2 When n ∈ N.
17. Show that all the roots of the equation a1z3 + a2z2 + a3z + a4 = 3, where |ai| ≤ 1, i = 1, 2, 3, 4 lie outside thecircle with centre origin and radius 2/3.
18. Prove that ∑−
=
−1n
1k
)kn( cos nk2 π
= – 2n
, where n ≥ 3 is an integer
19. Show that the equationA
x aA
x aA
x an
n
12
1
22
2
2
−+
−+ +
−...... = k has no imaginary root, given that :
a1, a2, a3.... an & A1, A2, A3..... An, k are all real numbers.20. Let z1, z2, z3 be three distinct complex numbers satisfying, ½z1-1½ = ½z2-1½ = ½z3-1½. Let A, B & C
be the points represented in the Argand plane corresponding to z1, z2 and z3 resp. Prove that z1 + z2 +z3 = 3 if and only if D ABC is an equilateral triangle.
21. Let α , β be fixed complex numbers and z is a variable complex number such that,
z − α 2 + z − β 2 = k.Find out the limits for 'k' such that the locus of z is a circle. Find also the centre and radius of thecircle.
22. If 1, α1, α2, α3,......., αn − 1 are the n, nth roots of unity, then prove that(1 − α1) (1 − α2) (1 − α3)........ (1 − αn − 1) = n.
Hence prove that sinπn
. sin2 πn
. sin3 πn
........ sin( )n
n− 1 π
=n
n2 1− .
23. Find the real values of the parameter ‘a’ for which at least one complex numberz = x + iy satisfies both the equality z − ai = a + 4 and the inequality z − 2 < 1.
24. Prove that, with regard to the quadratic equation z2 + (p + ip′) z + q + iq′ = 0; where p, p′ , q, q′ are allreal.(a) if the equation has one real root then q ′2 − pp ′ q ′ + qp ′2 = 0.(b) if the equation has two equal roots then p2 − p′2 = 4q & pp ′ = 2q ′ .
State whether these equal roots are real or complex.25. The points A, B, C depict the complex numbers z1, z2, z3 respectively on a complex plane & the angle
B & C of the triangle ABC are each equal to12
( )π α− . Show that
(z2 − z3)² = 4 (z3 − z1) (z1 − z2) sin2 α2
.
26. If z1, z2 & z3 are the affixes of three points A, B & C respectively and satisfy the condition|z1 – z2| = |z1| + |z2| and |(2 - i) z1 + iz3 | = |z1| + |(1 – i) z1 + iz3| then prove that ∆ ABC in a right angled.
27. If 1, α1, α2, α3, α4 be the roots of x5 − 1 = 0, then prove that
12
1
α−ω
α−ω .2
22
α−ω
α−ω .3
23
α−ω
α−ω .4
24
α−ω
α−ω = ω.
28. If one the vertices of the square circumscribing the circle |z – 1| = 2 is 2 + 3 i. Find the other vertices ofthe square. [IIT – 2005, 4]
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29 of
38
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
FREE
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nloa
d St
udy
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age
Vie
ws o
f stu
dent
s ava
ilabl
e at
web
site:
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w.ii
tjeei
itjee
.com
, w
ww
.teko
clas
ses.c
om
EXEREXEREXEREXEREXERCISE-4CISE-4CISE-4CISE-4CISE-41. A 2. C 3. D 4. A
5. A 6. D 7. C 8. A
9. A 11. D 12. A 13. B
14. D 15. D 16. A 17. B
18. A 19. D 20. C 21. C
22. B 23. B 24. D 25. B
26. C 27. A 28. ABCD29. AB
30. ACD 31. AC 10. AD
EXEREXEREXEREXEREXERCISE-5CISE-5CISE-5CISE-5CISE-51. x = K, y =
32K
K ∈ R 3. 32,32 −−
4. 3 – 2i
or 1 – 23
i 11. 3 ω2
21. k > 12
2α β− 23. − −
2110
56
,
28. – i 3 , 1 – 3 + i, 1 + 3 – i