i.b. mathematics hl core: complex numbers question 1 ... · pdf filei.b. mathematics hl core:...
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I.B. Mathematics HL Core: Complex Numbers Index: Please click on the question number you want
Question 1 Question 2 Question 3 Question 4
Question 5 Question 6
Question 7 Question 8 Question 9 Question 10
Question 11
You can access the solutions from the end of each question
Question 1 For 3 4w i= + and 5 12z i= − Find: a. w z+ b. w z− c. wz
d. wz
Give your answers in the form a bi+ , where ,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 1 a. 3 4 5 1
8 82w i i
iz+ = −
= −+ +
b. ( )
2 16
3 4 5 123 4 5 12
w z i iI I
i
+ = + − −
= +− +
− +=
c. ( )( )
( ) ( )2 2
3 4 5 12
3 5 12 4 5 12
15 36 20 4 Remember8 15 36 20 4
3 1
1
68
6
wz i i
i i i
i ii
ii
ii
=
= + −
= − + −
= − + −= − + +=
−
−
d.
( ) ( )2
2
2
3 45 123 5 1
33 56169 16
2 4 5 1225 144
15 36 20 4825 144
multply by complex conjugat5 125 12
Remembe
e
r 1
9
w iz i
i i ii
i i i
ii
i
i
+=−
+ + +
− +
++
= −=−
+ + +=+
=
Click here to read the question again Click here to return to the index
Question 2
Given that 2 3z i= + and 11
wi
=+
, write down the exact values of Re(w)
and Im(zw). Click here to read the solution to this question Click here to return to the index
Solution to question 2
( ) 21 1 1 1Re Re Re Re Re
1 1 1 2112
1 i iwi i
ii i
− − = = = = = + + − −−
( ) ( ) ( )
( )2
2
2 1 3 12 3Im Im Im1 1
2 3 3 22 2 3 3Im Im1 1
1
32
1
2
2
i ii iizwi
i i
ii
ii
− + −+= = + − + + − − + − = = +
−
−
−
=
( )2 1i = −
Click here to read the question again Click here to return to the index
Question 3 Find ( )4
3 3i− + in the form a bi+ , where ,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 3 First write 3 3i− + into polar form ( )cos sinr iθ θ+ .
Therefore 3 3 2 3 cosi − + =
( ) (43 3 2i⇒ − + =
Using De Moivre’s theorem we
( )43 3 144 c
144 c
14
2 27 7
4
i − + = =
= −−=
Click here to read the questi Click here to return to the in
r
θ
3
3
Re(z)
Im(z)
( ) ( )223 3 12 2 3r = − + = =5 5sin6 6
iπ π +
)4
4 5 53 cos sin6 6
iπ π +
have:
5 5os4 sin46 6
10 10os sin3 3
1 32
3
2
i
i
i
iπ π
π π
+ +
−
on again
dex
( ) 3 5arctan3 6 6
Arg z π πθ π π
= = − = − =
Question 4 Find 5 12i− + giving your answers in the form a bi+ , where ,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 4 Let 5 12z i= − + Now z a bi= + ( )22 5 12z i a bi⇒ = − + = +
( )2
2
2
2
2 25 12 2
12
1
5 2
i a abi b i
b
i
i aa bi
=− + = + +
+ = +
−
− −
Equating the real and imaginary parts we have
2 25 a b− = − and 12 62ab ba
= ⇒ = 2
2
22
2 4
4 2
65
365
5 365 36 0
aa
aa
a aa a
− = −
− = −
− = −+ − =
product= -36
sum = 5 factors = -4, 9
( ) ( )( )( )
4
2 2 2
2
2 2
2
36 04 9 4
9
90
4
4
0
a
a a a
a a
a a − =
− + − =
+ −
+
=
−
2a⇒ = ± as a is real.
When 2 3a b= − ⇒ = − and when 2 3a b= ⇒ = Giving and 2 3 2 3z i z i= − − = + Click here to read the question again Click here to return to the index
Question 5 Find the cube root of 2 2i+ , giving exact answers in the form a bi+ , where
,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 5 To solve 3 2 2z i= + First write 2 2i+ into polar form ( )cos sinr iθ θ+ .
( )1 13 31
63 32 2 cos sin 2 2 cos sin 2 cos sin4 4 4 4 4 4
z i z i iπ π π π π π = + ⇒ = + = + 1 12 cos 2 sin 2 for 0,1, 2.3 4 3 4
z k i k kπ ππ π ⇒ = + + + =
0For 0 2 cos sin12 12
k z iπ π = = +
Fo
Cl
2 ( )
2 22 2 8 2 22arctan2 4
r
Arg z πθ
= + = =
= = =
r
θ
2 Re(z)
Im(z)
Now 1 2 3 2cos cos cos cos sin sin12 3 4 3 4 3 4 2 2 2 2π π π π π π π = − = + = +
( )2 2 3 2 1 34 4 4
= + = +
Now 3 2 1 2sin sin sin cos cos sin12 3 4 3 4 3 4 2 2 2 2π π π π π π π = − = − = −
2 3 2 2
( ) ( )02 22 1 3 3 1
4 41 3 3 1
2 2iz i
⇒ + −+= + + − =
03 3 2 2r 1 2 cos sin 24 4 2 2
1k z i iiπ π = = + = − + = −
+
ick here to continue with solution or go to next page
( )3 14 4 4
= − = −
Continue with solution to question 5
217 17For 2 2 cos sin12 12
k z iπ π = = +
( ) ( )22 22 1 3 3 1
4 41 3 3 1
2 2iz i
⇒ − +−= − − + =
1 3 3 1 1 3 3 1, 1 or 2 2 2 2
z i z i z i+ − − += + = − + = −
Click here to read the question again Click here to return to the index
Now 17 5 5 5 1 2 3 2cos cos cos cos sin sin12 3 4 3 4 3 4 2 2 2 2
π π π π π π π = − = + = + −
( )2 2 3 2 1 34 4 4
= − = −
Now 17 5 5 5 3 2 1 2sin sin sin cos cos sin12 3 4 3 4 3 4 2 2 2 2
π π π π π π π = − = − = − −
( )2 3 2 2 3 14 4 4
= − − = − +
Question 6 Solve the equation 25 4 3 0z z+ + = , giving exact answers in the form a bi+ ,
,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 6 Now 25 4 3 0z z+ + =
Solving using the quadratic formula 2 4
2b b acz
a− ± −=
( )( )( )
2
24 4 4 5 32 5
4 16 6010
4 44 110
4 4410
4 2 1110
2 115
1 1i i
z
i
i
i
=
− ± −=
− ± −=
− ± −=
− ±=
− ±=
− ±=
− ⇒ = −
Giving or 2 11 2 115 5 5 5
z i z i= − − = − +
(Notice that the solutions occur in conjugate pairs for real coefficients). Click here to read the question again Click here to return to the index
Question 7 Given that 2 3i+ is a solution of the equation 4 3 22 3 3 77 39 0z z z z− + + − = , find the other solutions. Click here to read the solution to this question Click here to return to the index
Solution to question 7 If 2 3z i= + is a solution of 4 3 22 3 3 77 39 0z z z z− + + − = then 2 3z i= − is also a solution as complex roots occur in conjugate pairs for polynomials with real coefficients.
( )( )2 3 2 3z i z i⇒ − − − + must be factors of 4 3 22 3 3 77 39z z z z− + + − . ⇒ ( )( ) ( ) ( ) ( )
2 22
2
2 3 2 3 2 3 2 2 3 3 2 3
2 3 2 4 6 3 6 94 13
1
z z i z i z z i z i i z i
z z iz z i iz i iz z
i
− − − + = − + − − + − − +
= − + − + − − + −= − +
= −
Hence: 4
3
2
2 3 2
4 3 2
2
3 2
2
2
2
5
2 5 34 13 3 3 77 39
2 8 2623 72
5 20 6512 39
3 393
12
z zz z z z z
z z zz z
z z zz
z
z
zz
z+ −
− + − − + −
− +− +
+ −− +−
− +
−
Now solving: 22 5 3 0z z+ − =
Product = -6 Sum = 5 Factors = -1, 6
( ) ( )( )( )
22 3 02 1 3 2 1 0
2 1 3 0
6zzz z z
z z
z − =− + − =
− +
+
=
−
Therefore the other solutions are 2 3z i= − , 3z = − or 12
z =
Click here to read the question again Click here to return to the index
Question 8 If z x iy= + , where ,x y ∈ and 1i = − find the Cartesian equation of
2 3z z i− = − and represent it on an Argand diagram. Click here to read the solution to this question Click here to return to the index
Solution to question 8 If 2 3z z i− = − then:
( )2 3
2 3
x iy x iy i
x iy x i y
+ − = + −
− + = + −
as 2 2z x iy x y= + = + and squaring both sides, we have ( ) ( )2 22 2
2 2 2 2
2 3
4 44
96
65 0
x y x y
x x y xx y
y y
− + = + −
− + +−
− ++
+==
Argand diagram Click here to read the question aga Click here to return to the index
)
3i
2
4 6 5 0x y− + =
Im(zin
Re(z)
Question 9 If z x iy= + , where ,x y ∈ and 1i = − find the Cartesian equation of
( )2 34
Arg z i π− + = and represent it on an Argand diagram.
Click here to read the solution to this question Click here to return to the index
Solution of question 9
( ) ( )( )2 3 2 34 4
Arg z i Arg z iπ π− − = ⇒ − + =
Now 3 tan2 4
yx
π− =−
⇒ 3 123 2
1
yxy x
y x
− =−− = −
= +
Therefore the Cartesian equation is 1 for 2y x x= + ≥ (A half line). Argand diagram Click here to read the question again Click here to return to the index
)
3i
2
1y x= +
Im(z Re(z)
Question 10 If z x iy= + , where ,x y ∈ and 1i = − find the Cartesian equation of
1 2z z i+ = − and represent it on an Argand diagram. Click here to read the solution to this question Click here to return to the index
Solution to question 10 If 1 2z z i+ = − then:
( )1
1 1
x iy x iy i
x iy x i y
+ + = + −
− + = + −
as 2 2z x iy x y= + = + and squaring both sides, we have ( ) ( )( )
( )
2 22 2
2 2 2 2
2 2 2 2
2 2
2 2
1 4 1
2 1 4 2 1
2 1 4 4 8 43 2 3 8 0
2 8 03 3
x y x y
x x y x y y
x x y x y yx x y y
x x y y
+ + = + −
+ + + = + − +
+ + + = + − +− + − =
− + − =
Completing the square
2
2
2
2
1 4 1 16
1 4 173
3 3 9
3 9
9
x y
x y
− + + =
− + + = +
Which is a circle centre 1 4,3 3
− , radius 17
3
Argand diagram Click here to read the Click here to return to
43
i−
)
ques
the i
13
)
Im(z
tion ag
ndex
213
x −
Re(z
ain
24 173 3
y + − =
Question 11 Solve the equation 4 81z i= giving exact answers in the form a bi+ , where
,a b ∈ . Click here to read the solution to this question Click here to return to the index
Solution to question 11 To solve 4 81z i= First write 81i into polar form ( )cos sinr iθ θ+ .
4 81 cos sin2 2
z i zπ π = + ⇒ = 13 cos 2 sin4 2
z k iπ π ⇒ = + +
For 0k = 0 3 cos sin8 8
z iπ π= +
2 2
2
2
cos2 cos sin
2cos 1 co
1 2sin sin
θ θ θ
θ
θ
= −
= − ⇒
= − ⇒
8π is in the 1st quadrant theref
( )
( )
1cos 1 cos2 c2
1sin 1 cos2 s2
θ θ
θ θ
= ± + ⇒
= ± − ⇒
(0 3 cos sin8
3 228
z iπ π = + = Click here to continue with s
81 r
θ
Re(z)
Im(z)
1 14 41
481 cos sin 3 cos sin2 2 2 2
i iπ π π π + = + 1 2 for 0,1, 2, 3.4 2
k kπ π + =
( ) ( )
( ) ( )
2
2
1cos 1 cos22
1s
1s 1 cos22
1 1 cos2 in 1 cos22
2
θ θθ θ
θ θ θ θ
= + ⇒
= − ⇒
= ± +
= ± −
ore both cosine and sine are both positive.
1 1 2 1os 1 cos 1 2 28 2 4 2 2 2
1 1 2 1in 1 cos 1 2 28 2 4 2 2 2
π π
π π
= + = + = +
= − = − = −
)2 2 2i+ + −
olution or go to next page
( )
2 2 20 81 81 81
2
r
Arg z πθ
= + = =
= =
Continue with solution to question 11
For 1k = 15 53 cos sin8 8
z iπ π = +
58π is in the 2nd quadrant therefore cosine is negative and sine is positive.
( )
( )
1 5 1 5 1 2 1cos 1 cos2 cos 1 cos 1 2 22 8 2 4 2 2 2
1 5 1 5 1 2 1sin 1 cos2 sin 1 cos 1 2 22 8 2 4 2 2 2
π πθ θ
π πθ θ
= ± + ⇒ = − + = − − = − −
= ± − ⇒ = − = + = +
( )15 53 cos sin8
3 2 2 2 228
z i iπ π = + = −
−
+ −
For 2k = 29 93 cos sin8 8
z iπ π = +
98π is in the 3rd quadrant therefore cosine is negative and sine is negative.
( )
( )
1 9 1 9 1 2 1cos 1 cos2 cos 1 cos 1 2 22 8 2 4 2 2 2
1 9 1 9 1 2 1sin 1 cos2 sin 1 cos 1 2 22 8 2 4 2 2 2
π πθ θ
π πθ θ
= ± + ⇒ = − + = − + = − +
= ± − ⇒ = − − = − − = − −
( )29 93 cos sin8
3 2 2 2 228
z i iπ π = + = −
+
+ −
For 3k = 213 133 cos sin
8 8z iπ π = +
138π is in the 4th quadrant therefore cosine is positive and sine is negative.
( )
( )
1 13 1 13 1 2 1cos 1 cos2 cos 1 cos 1 2 22 8 2 4 2 2 2
1 13 1 13 1 2 1sin 1 cos2 sin 1 cos 1 2 22 8 2 4 2 2 2
π πθ θ
π πθ θ
= ± + ⇒ = + = − = −
= ± − ⇒ = − − = − + = − +
( )213 133 cos sin
83 2 2 2 228
z i iπ π = + = − − +
Click here to continue with solution or go to next page
Continue with solution to question 11 The solutions are:
( )03 2 2 2 22
z i= + + − , ( )13 2 2 2 22
z i= − − − + ,
( )23 2 2 2 22
z i− − + + and ( )33 2 2 2 22
z i= − − + .
Notice when the solutions are plotted on an Argand diagram they are evenly spaced and form a circle radius 3. Click here to read the que Click here to return to the
Re(z)
Im(z)
98π
58π
8π
z0
z1
z2
stion again
index
138π
z3
3 units