hws

United Arab Emirates University

College of Sciences

Department of Mathematical Sciences

HOMEWORK 1 { SOLUTION

Section 1 { Section 5

Complex Analysis I

MATH 315 SECTION 01 CRN 23516

9:30 { 10:45 on Monday & Wednesday

Due Date: Monday, September 13, 2009

Complex Analysis I HOMEWORK 1 { SOLUTION Fall, 2009

Section 1. Sums and Products and Section 2 Basic Algebraic Properties

1. Verify that

(1.1)�p

2� i�� i

�1� i

p2�= �2i

Proof. p2� i� i+ i2

p2 =

p2� i� i+ (�1)

p2 = �2i: �

(1.2) (2;�3)(�2; 1) = (�1; 8)Proof.

(2;�3)(�2; 1) = (�4 + 3; 2 + 6) = (�1; 8): �

(1.3) (3; 1)(3;�1)�1

5;1

10

�= (2; 1)

Proof.

(3; 1)(3;�1) = (9 + 1;�3 + 3) = (10; 0);

(3; 1)(3;�1)�1

5;1

10

�= (10; 0)

�1

5;1

10

�= (2� 0; 1 + 0) = (2; 1): �

2. Show that

(2.1) Re(iz) = � Im(z)

Proof. For z = x+ yi,

iz = i(x+ yi) = �y + xi; Re(iz) = �y = � Im(z); Re(iz) = � Im(z): �

(2.2) Im(iz) = Re(z)

Proof. For z = x+ yi,

iz = i(x+ yi) = �y + xi; Im(iz) = x = Re(z); Im(iz) = Re(z): �

Page 1 of 7


3. Solve the equation z2 + z + 1 = 0 for z = (x; y) by writing

(x; y)(x; y) + (x; y) + (1; 0) = (0; 0)

and then solving a pair of simultaneous equations in x and y.

Answer. We expand the equation,

(0; 0) = (x; y)(x; y) + (x; y) + (1; 0)

= (x2 � y2; 2xy) + (x; y) + (1; 0) = (x2 � y2 + x+ 1; 2xy + y + 0);

i:e:; 0 = x2 � y2 + x+ 1; and 0 = 2xy + y = y(2x+ 1): (1)

From the second equation in (1), we get y = 0 or 2x+ 1 = 0, i.e., x = �1=2.Case 1. y = 0: Putting y = 0 into the �rst equation in (1), we get

0 = x2 + x+ 1;

which cannot be solved for a real x. That is, y = 0 cannot be the solution of the given equation.

Case 2. x = �1=2: Putting x = �1=2 into the �rst equation in (1), we get

0 =1

4� y2 � 1

2+ 1 =

3

4� y2; y = �

p3

2:

It implies the solutions to the given equation z2 + z + 1 = 0 are

z = (x; y) =

�1

2;�p3

2

!: �

Section 3. Further Properties

4. Reduce each of these quantities to a real number:

(4.1)1 + 2i

3� 4i+

2� i

5i

Answer.1 + 2i

3� 4i+

2� i

5i=

1 + 2i

3� 4i

3 + 4i

3 + 4i

!+

2� i

5i

�5i�5i

!= �2

5: �

(4.2)5i

(1� i)(2� i)(3� i)

Answer.5i

(1� i)(2� i)(3� i)=

5i

�10i = �1

2: �

(4.3) (1� i)4

Answer.

(1� i)4 =h(1� i)2

i2

= [�2i]2 = �4: �

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5. Prove that if z1z2z3 = 0, then at least one of the three factors is zero.

Proof. We recall that for complex numbers z and w, if zw = 0, then z = 0 or w = 0 or possiblyboth z and w equal zero. It implies that if (z1z2)z3 = z1z2z3 = 0, then z1z2 = 0 or z3 = 0 orpossibly both z1z2 and z3 equal zero. Again if z1z2 = 0, then z1 = 0 or z2 = 0 or possibly bothz1 and z2 equal zero. Therefore, combining those two statements, we can deduce that z1 = 0 orz2 = 0 or z3 = 0, i.e., at least one of the three factors is zero. �

6. Use identityz1z2z3z4

=�z1z3

��z2z4

�; z3 6= 0 6= 0z4;

to derive the cancelation law:z1z

z2z=

z1z2; z2 6= 0 6= z:

Proof. By the identity,

z1z

z2z=�z1z2

��z

z

�=

z1z2

�zz�1

�=

z1z2; z2 6= 0 6= z: �

Section 4. Moduli

7. Locate the numbers z1 + z2 and z1 � z2 vectorially when

(7.1) z1 = 2i and z2 =2

3� i

Answer. z1 = 2i = (0; 2), z2 = 2=3� i = (2=3;�1).

z1 + z2 = (2=3; 1); z1 � z2 = (�2=3; 3): �

(7.2) z1 = (�p3; 1) and z2 = (

p3; 0)

Answer. z1 = (�p3; 1), z2 = (p3; 0).

z1 + z2 = (0; 1); z1 � z2 = (�2p3; 1): �

(7.3) z1 = (�3; 1) and z2 = (1; 4)

Answer. z1 = (�3; 1), z2 = (1; 4).

z1 + z2 = (�2; 5); z1 � z2 = (�4;�3): �

(7.4) z1 = x1 + y1i and z2 = x1 � y1i

Answer. z1 = x1 + y1i = (x1; y1), z2 = x1 � y1i = (x1;�y1).

z1 + z2 = (2x1; 0); z1 � z2 = (0; 2y1): �

Page 3 of 7


8. Verify that p2jzj � jRe(z)j+ j Im(z)j:

Proof. Let z = x+ yi. Then

Re(z) = x; and Im(z) = y; and jzj2 = x2 + y2 = jxj2 + jyj2:

We observe

0 � (jxj � jyj)2 = jxj2 + jyj2 � 2jxjjyj; i:e:; 2jxjjyj � jxj2 + jyj2:

Adding jxj2 + jyj2 to both sides, we get

jxj2 + jyj2 + 2jxjjyj � 2�jxj2 + jyj2

�; i:e:; (jxj+ jyj)2 � 2

�jxj2 + jyj2

�:

Now we take both sides to be squared. The inequality does not change the order, because bothsides are nonnegative.

q(jxj+ jyj)2 �

p2q(jxj2 + jyj2); i:e:; jxj+ jyj �

p2q(jxj2 + jyj2);

which is jRe(z)j+ j Im(z)j �p2jzj. �

9. In each case, sketch the set of points determined by the given condition:

(9.1) jz � 1 + ij = 1

Answer. It is a circle centered at the point 1� i with radius 1. �(9.2) jz + ij � 3

Answer. It is a disc inside and including the circle centered at the point �i with radius3. �

(9.3) jz � 4ij � 4

Answer. It is a region outside and including the circle centered at the point 4i with radius4. �

Section 5. Complex Conjugates

10. Use properties of conjugates and moduli to show that

(10.1) �z + 3i = z � 3i

Proof.

�z + 3i = ��z + 3i = z � 3i: �

(10.2) iz = �i�zProof.

iz = �i�z = �i�z: �

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(10.3) (2 + i)2 = 3� 4i

Proof. Since (2 + i)2 = 3 + 4i, so

(2 + i)2 = 3 + 4i = 3� 4i: �

(10.4) j (2�z + 5)�p

2� i�j =

p3j2z + 5j

Proof. Recall j�zj = jzj.

��(2�z + 5)�p

2� i�� = j2�z + 5j

��p2� i�� = p

3j2�z + 5j

=p3j2�z + 5j =

p3j2��z + 5j =

p3j2z + 5j: �

11. Sketch the set of points determined by the condition

(11.1) Re(�z � i) = 2

Answer. We recall Re(z) = Re(�z). So

Re (�z � i) = Re��z � i

�= Re

��z � �i

�= Re (z + i) :

That is, the given equation is equivalent to Re (z + i) = 2. Let z = x+ yi and put into theequation, then

x = Re(x+ (y + 1)i) = Re(x+ yi+ i) = Re(z + i) = 2; i:e:; x = 2:

It implies that for any y, z = 2 + yi satis�es the equation. That is, the line x = 2 in thecomplex plane satis�es the given equation. �

(11.2) j2z � ij = 4

Answer. We simplify the given equation

2

��z � i

2

�� = 4; i:e:;

��z � i

2

�� = 2;

which represents a circle centered at the point i=2 with radius 2. �

Page 5 of 7


12. Use established properties of moduli to show that when jz3j 6= jz4j,��z1 + z2z3 + z4

�� jz1j+ jz2jjjz3j � jz4jj :

Proof. The Triangle Inequality implies

jz1 + z2j � jz1j+ jz2j; and so��z1 + z2z3 + z4

�� = jz1 + z2jjz3 + z4j �

jz1j+ jz2jjz3 + z4j :

Now we recall, from the triangle inequality,

jjz3j � jz4jj � jz3 + z4j; and so 0 <1

jz3 + z4j �1

jjz3j � jz4jj :

Therefore, we conclude

��z1 + z2z3 + z4

�� jz1j+ jz2jjz3 + z4j �

jz1j+ jz2jjjz3j � jz4jj ; i:e:;

��z1 + z2z3 + z4

�� jz1j+ jz2jjjz3j � jz4jj : �

13. Show that ��Re(2 + �z + z3)�� 4; when jzj � 1:

Proof. Recall for a complex number w, jRe(w)j � jwj. So for jzj � 1, we have

��Re(2 + �z + z3)�� 2 + �z + z3

�� j2j+ j�zj+ jz3j = 2 + jzj+ jzj3 � 2 + 1 + 1 = 4: �

14. Prove that

(14.1) z is real if and only if �z = z.

Proof. Let z = x+ yi.

If z is real, then y = 0 and so z = x, i.e., �z = z.

If �z = z, then x� yi = x+ yi, 2yi = 0, i.e., y = 0 and so z = x+ yi becomes a real x. �

(14.2) z is either real or pure imaginary if and only if �z2 = z2.

Proof. Let z = x+ yi.

If z is real, i.e., y = 0, then from the problem above, �z = z and so �z2 = z2.

If z is pure imaginary, i.e., x = 0 and so z = yi, then �z = �yi and �z2 = �y2 = z2, i.e., �z2 = z2.

If �z2 = z2, then �z2 � z2 = 0, i.e.,

0 = �z2 � z2 = (x� yi)2 � (x+ yi)2 = x2 � y2 � 2xyi� (x2 � y2 + 2xyi) = �4xyi;

i.e., y = 0 or x = 0, which means z is either real or pure imaginary. �

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15. Using expressions,

Re(z) =z + �z

2; Im(z) = �z � �z

2i;

show that the hyperbola x2 � y2 = 1 can be written

z2 + �z2 = 2:

Proof. Let z = x+ yi. Then by the given expressions,

x =z + �z

2; y = �z � �z

2i:

Putting them into the hyperbola equation x2 � y2 = 1, we have

1 =�z + �z

2

�2

��z � �z

2i

�2

=(z + �z)2

4+

(z � �z)2

4

=1

4

�z2 + �z2 + 2z�z + z2 + �z2 � 2z�z

�=

1

2

�z2 + �z2

�; i:e:; z2 + �z2 = 2: �

Page 7 of 7

United Arab Emirates UniversityCollege of Sciences


HOMEWORK 2 – SOLUTION

Complex Analysis IMATH 315 SECTION 01 CRN 235169:30 – 10:45 on Monday & WednesdayDue Date: Monday, September 28, 2009

Complex Analysis I HOMEWORK 2 – SOLUTION Fall, 2009

Section 6. Exponential FormSection 7. Products and Quotients in Exponential Form

1. Find the principal argument Arg(z) when

(1.1) z =i

−2− 2i

Answer.

z =i

−2− 2i= −1

4(1 + i) , tan (Arg(z)) =

−1/4

−1/4= 1, Arg(z) = −3π

4. �

(1.2) z =(√

3− i)6

Answer. Let w =√

3− i. Then z = w6 and

tan (Arg(w)) =−1√

3= −π

6, Arg(w) = −π

6, Arg(z) = Arg(w6) = 6 Arg(w) = −π,

where Arg(z1z2) = Arg(z1) + Arg(z2) is used. Since −π < Arg(z) ≤ π, hence we conclude

Arg(z) = π.

In fact, z =(√

3− i)6

= −64 and so Arg(z) = π. �

2. Show that

(2.1) |eiθ| = 1

Proof. By Euler’s formula,

eiθ = cos θ + i sin θ, |eiθ| = | cos θ + i sin θ| =√

cos2 θ + sin2 θ = 1. �

(2.2) eiθ = e−iθ .

Proof. By Euler’s formula,

eiθ = cos θ + i sin θ = cos θ − i sin θ = cos(−θ) + i sin(−θ) = e−iθ. �

3. Using the fact that the modulus∣∣eiθ − 1

∣∣ is the distance between the points eiθ and 1, give ageometric argument to find a value of θ in the interval 0 ≤ θ < 2π that satisfies the equation∣∣eiθ − 1

∣∣ = 2.

Answer. Since eiθ, 0 ≤ θ < 2π, has the modulus |eiθ| = 1, so the points eiθ, 0 ≤ θ < 2π, are allon the unit circle centered at the origin. (The unit circle means a circle with radius 1.) If eiθ,0 ≤ θ < 2π, satisfies the equation |eiθ − 1| = 2, then the point eiθ is 2 units away from the point(1, 0) in the plane. Hence the point eiθ should be at (−1, 0) and it means θ = π. �

4. Use de Movire’s formula to derive the following trigonometric identities:

(4.1) cos(3θ) = cos3 θ − 3 cos θ sin2 θ

(4.2) sin(3θ) = 3 cos2 θ sin θ − sin3 θ

Page 1 of 9


Proof.

eiθ = cos θ + i sin θ,

cos(3θ) + i sin(3θ) = ei3θ = (eiθ)3 = (cos θ + i sin θ)3

= cos3 θ + 3i cos2 θ sin θ − 3 cos θ sin2 θ − i sin3 θ

= cos3 θ − 3 cos θ sin2 θ + i(3 cos2 θ sin θ − sin3 θ

).

By equating the real parts and then the imaginary parts, we arrive at the desired trigonometricidentities,

cos(3θ) = cos3 θ − 3 cos θ sin2 θ, sin(3θ) = 3 cos2 θ sin θ − sin3 θ. �

5. By writing the individual factors on the left in exponential form, performing the needed operators,and finally changing back to rectangular coordinates, show that

(5.1) i(1− i√

3)(√

3 + i) = 2(1 + i√

3)

Answer. It is straightforward to see i = eπi2 , 1− i

√3 = 2e−

πi3 ,√

3 + i = 2eπi6 . So,

i(1− i√

3)(√

3 + i) = eπi2 2e−

πi3 2e

πi6 = 4e

πi3 = 4

(cos

π

3+ i sin

π

3

)= 2

(1 + i

√3). �

(5.2)5i

2 + i= 1 + 2i

Answer. It is straightforward to see 5i = 5eπi2 , 2 + i =

√5ei tan

−1 12 . So,

5i

2 + i=

5eπi2

√5ei tan

−1 12

=√

5ei(π2−tan−1 1

2) =√

5

[cos

(π

2− tan−1

1

2

)+ i sin

(π

2− tan−1

1

2

)]=√

5

[1√5

+ i2√5

]= 1 + 2i. �

(5.3) (−1 + i)7 = −8 (1 + i)

Answer. It is straightforward to see −1 + i =√

2e3πi4 . So,

(−1 + i)7 =(√

2)7(

e3πi4

)7

= 27/2e21πi4 = 27/2

(cos

21π

4+ i sin

21π

4

)= 27/2

(− 1√

2− i 1√

2

)= −8 (1 + i) . �

(5.4)(

1 + i√

3)−10

= 2−11(−1 + i

√3)

Answer. It is straightforward to see 1 + i√

3 = 2eπi3 . So,

(1 + i√

3)−10 = 2−10(eπi3

)−10= 2−10e−

10πi3 = 2−10

(cos

10π

3− i sin

10π

3

)

Page 2 of 9


= 2−10

(−1

2+ i

√3

2

)= 2−11

(−1 + i

√3). �

6. Prove that two nonzero complex numbers z1 and z2 have the same moduli if and only if there arecomplex numbers c1 and c2 such that z1 = c1c2 and z2 = c1c̄2.

Proof. Suppose there are complex numbers c1 and c2 such that z1 = c1c2 and z2 = c1c̄2. Then, bythe properties on moduli, we have

|z1| = |c1c2| = |c1||c2| = |c1||c̄2| = |c1c̄2| = |z2|, i.e., |z1| = |z2|.

Suppose that two nonzero complex numbers z1 and z2 have the same moduli. Let |z1| = r1 = |z2|and Arg(z1) = r1 and Arg(z2) = θ2. Then we may write

z1 = r1eiθ1 , z2 = r2e

iθ2 .

If we introduce the numbers

c1 = r1eiθ1+θ2

2 , and c2 = eiθ1−θ2

2 ,

we find that

c1c2 = r1eiθ1+θ2

2 eiθ1−θ2

2 = r1eiθ1 = z1, and c1c̄2 = r1e

iθ1+θ2

2 e−iθ1−θ2

2 = r1eiθ2 = z2.

That is,z1 = c1c2, and z2 = c1c̄2.

Hence, there are complex numbers c1 and c2 desired in the problem. �

7. Establish the identity

1 + z + z2 + · · ·+ zn =1− zn+1

1− z, z 6= 1,

and then use it to derive Lagrange’s trigonometric identity:

1 + cos θ + cos(2θ) + · · ·+ cos(nθ) =1

2+

sin[(2n+ 1)θ/2]

2 sin(θ/2), 0 < θ < 2π.

Proof. Let S =∑n

k=0 zk. Then we observe for z 6= 1,

(1− z)S = S − zS =n∑k=0

zk −n∑k=0

zk+1 =n∑k=0

(zk − zk+1

)= 1− zn+1, S =

1− zn+1

1− z.

Substituting z = eiθ into the equation above, we get

n∑k=0

eikθ =1− ei(n+1)θ

1− eiθ.

Page 3 of 9


Euler’s formula implies

n∑k=0

cos(kθ) + in∑k=0

sin(kθ) =n∑k=0


1− eiθ

=1− ei(n+1)θ

1− eiθ

(e−iθ/2

e−iθ/2

)=e−iθ/2 − ei(2n+1)θ/2

e−iθ/2 − eiθ/2,

Again Euler’s formula implies

n∑k=0

cos(kθ) + in∑k=0

sin(kθ) =e−iθ/2 − ei(2n+1)θ/2

e−iθ/2 − eiθ/2

=cos(θ/2)− i sin(θ/2)− [cos((2n+ 1)θ/2) + i sin((2n+ 1)θ/2)]

−2i sin(θ/2)

=cos(θ/2)− i sin(θ/2)− [cos((2n+ 1)θ/2) + i sin((2n+ 1)θ/2)]

−2i sin(θ/2)

i

i

=sin(θ/2) + sin((2n+ 1)θ/2) + i [cos(θ/2)− cos((2n+ 1)θ/2)]

2 sin(θ/2).

By equating the real parts and the imaginary parts, we find

n∑k=0

cos(kθ) =sin(θ/2) + sin((2n+ 1)θ/2)

2 sin(θ/2)=

1

2+

sin((2n+ 1)θ/2)

sin(θ/2),

n∑k=0

sin(kθ) =cos(θ/2)− cos((2n+ 1)θ/2)

2 sin(θ/2)=

1

2cot(θ/2)− cos((2n+ 1)θ/2)

sin(θ/2),

where 0 < θ < 2π. �

Another Proof. Euler’s formula implies

n∑k=0

cos(kθ) + in∑k=0

sin(kθ) =n∑k=0


1− eiθ

=1− ei(n+1)θ

1− eiθ

(1− e−iθ

1− e−iθ

)=

1− e−iθ + einθ − ei(n+1)θ

2− e−iθ − eiθ

=1− cos θ + cos(nθ)− cos(n+ 1)θ + i [sin θ + sin(nθ)− sin(n+ 1)θ]

2(1− cos θ)

That is, we deduce

n∑k=0

cos(kθ) =1− cos θ + cos(nθ)− cos(n+ 1)θ

2(1− cos θ)=

1

2+

cos(nθ)− cos(n+ 1)θ

2(1− cos θ),

n∑k=0

sin(kθ) =sin θ + sin(nθ)− sin(n+ 1)θ

2(1− cos θ),

where we will simplify the first one.

Page 4 of 9


We recall from Calculus

1− cosA = 2 sin2(A/2), sin(2A) = 2 sinA cosA,

cos(A+B) = cosA cosB − sinA sinB, sin(A+B) = sinA cosB + cosA sinB.

Using them, we have

n∑k=0

cos(kθ) =1

2+

cos(nθ)− cos(n+ 1)θ

2(1− cos θ)

=1

2+

cos(nθ)− cos(nθ) cos θ + sin(nθ) sin θ

4 sin2(θ/2)

=1

2+

cos(nθ) (1− cos θ) + 2 sin(nθ) sin(θ/2) cos(θ/2)

4 sin2(θ/2)

=1

2+

2 cos(nθ) sin2(θ/2) + 2 sin(nθ) sin(θ/2) cos(θ/2)

4 sin2(θ/2)

=1

2+

cos(nθ) sin(θ/2) + sin(nθ) cos(θ/2)

2 sin(θ/2)

=1

2+

sin(nθ + θ/2)

2 sin(θ/2)=

1

2+

sin[(2n+ 1)θ/2]

2 sin(θ/2). �

Page 5 of 9


Section 8. Roots of Complex Numbers and Section 9. Examples

8. Find the square roots of the followings and express them in rectangular coordinates.

(8.1) 2i

Answer. We look for z such that z2 = 2i, i.e., z = (2i)1/2. The complex number 2i has themodulus |2i| = 2 and the principal argument Arg(2i) = π/2. From the information, 2i hasthe exponential form,

2i = |2i|ei arg(2i) = 2ei(Arg(2i)+2kπ) = 2ei(π2+2kπ),

(2i)1/2 = 21/2ei(π4+kπ) =

√2ei(

π4+kπ), k = 0,±1,±2, . . . .

Hence, we deduce all two square roots by putting k = 0, 1,

c0 =√

2ei(π4 ) =

√2

(1√2

+1√2i

)= 1 + i,

c1 =√

2ei(π4+π) =

√2

(− 1√

2− 1√

2i

)= −1− i. �

(8.2) 1− i√

3

Answer. We look for z such that z2 = 1− i√

3, i.e., z = (1− i√

3)1/2. The complex number1 − i

√3 has the modulus |1 − i

√3| = 2 and the principal argument Arg(1 − i

√3) = −π/3.

From the information, 1− i√

3 has the exponential form,

1− i√

3 = |1− i√

3|ei arg(1−i√3) = 2ei(Arg(1−i

√3)+2kπ) = 2ei(−

π3+2kπ),

(1− i√

3)1/2 = 21/2ei(−π6+kπ) =

√2ei(−

π6+kπ), k = 0,±1,±2, . . . .

Hence, we deduce all two square roots by putting k = 0, 1,

c0 =√

2ei(−π6 ) =

√2

(√3

2− 1

2i

)=

√3√2− 1√

2i,

c1 =√

2ei(−π6+π) =

√2

(−√

3√2

+1√2i

)= −√

3√2

+1√2i. �

9. In each case, find all of the roots in rectangular coordinates, exhibit them as vertices of certainsquares, and point out which is the principal root.

(9.1) (−16)1/4

Answer. We look for z such that z = (−16)1/4. The complex number −16 has the modulus|−16| = 16 and the principal argument Arg(−16) = π. From the information, −16 has theexponential form,

−16 = |−16| ei arg(−16) = 16ei(Arg(−16)+2kπ) = 16ei(π+2kπ),

(−16)1/4 = 161/4e(2k+1)

4πi = 2e

(2k+1)4

πi, k = 0,±1,±2, . . . .

Page 6 of 9


Hence, we deduce all two square roots by putting k = 0, 1, 2, 3,

c0 = 2eπi4 = 2

(1√2

+1√2i

)=√

2 (1 + i) , c1 = 2e3πi4 = 2

(− 1√

2+

1√2i

)=√

2 (−1 + i) ,

c2 = 2e5πi4 = 2

(− 1√

2− 1√

2i

)= −√

2 (1 + i) , c3 = 2e7πi4 = 2

(1√2− 1√

2i

)=√

2 (1− i) .

The complex numbers c0, c1, c2 and c3 are respectively located at (√

2,√

2), (−√

2,√

2),(−√

2,−√

2), and (√

2,−√

2) in the complex plane and forms a square. The principal root isc0 located at (

√2,√

2). �(9.2) (−8− 8i

√3)1/4

Answer. We look for z such that z = (−8− 8i√

3)1/4. The complex number −8− 8i√

3 hasthe modulus

∣∣−8− 8i√

3∣∣ = 16 and the principal argument Arg(−8− 8i

√3) = −2π/3. From

the information, −8− 8i√

3 has the exponential form,

−8− 8i√

3 =∣∣∣−8− 8i

√3∣∣∣ ei arg(−8−8i√3) = 16ei(Arg(−8−8i

√3)+2kπ) = 16ei(−

2π3+2kπ),

(−8− 8i√

3)1/4 = 161/4ei(−π6+ kπ

2 ) = 2e(3k−1)

6πi, k = 0,±1,±2, . . . .

Hence, we deduce all two square roots by putting k = 0, 1, 2, 3,

c0 = 2e−πi6 = 2

(√3

2− 1

2i

)=√

3− i, c1 = 2eπi3 = 2

(1

2+

√3

2i

)= 1 + i

√3,

c2 = 2e5πi6 = 2

(−√

3

2+

1

2i

)= −√

3 + i, c3 = 2e4πi3 = 2

(−1

2−√

3

2i

)= −1− i

√3.

The complex numbers c0, c1, c2 and c3 are respectively located at (√

3,−1), (1,√

3), (−√

3, 1),and (−1,−

√3) in the complex plane and forms a square. The principal root is c0 located at

(√

3,−1). �

10. The three cube roots of a nonzero complex number z0 can be written c0, c0ω3, c0ω23, where c0 is

the principal cube root of z0 and

ω3 = exp

(2πi

3

)=−1 + i

√3

2.

Show that if z0 = −4√

2+4i√

2, then c0 =√

2(1+i) and the other two cube roots are in rectangularform, the numbers

c0ω3 =−(√

3 + 1) + (√

3− 1)i√2

, and c0ω23 =

(√

3− 1)− (√

3 + 1)i√2

.

Proof. We find all cube roots of z0 = −4√

2 (1− i). The given complex number z0 has the modulus|z0| =

√32 + 32 = 8 and the principal argument Arg(z0) = 3π/4. With this information, z0 has

the exponential form,

z0 = |z0|ei arg(z0) = |z0|ei(Arg(z0)+2kπ) = 8ei(3π4+2kπ)

Page 7 of 9


z1/30 = 81/3ei(

π4+ 2kπ

3 ) = 2eiπ4 ei

2kπ3 = 2ei

π4

[ei

2π3

]k= 2ei

π4ωk3 , k = 0,±1,±2, . . . .

Hence, we deduce all three cube roots by putting k = 0, 1, 2,

c0 = 2eiπ4ω0

3 = 2eiπ4 = 2

(1√2

+1√2i

)=√

2 (1 + i) ,

c1 = 2eiπ4ω1

3(= c0ω3) = 2eiπ4 e

2πi3 =

√2 (1 + i)

[−1 + i

√3

2

]=−(√

3 + 1) + (√

3− 1)i√2

,

c2 = 2eiπ4ω2

3(= c0ω23) = 2ei

π4

[e

2πi3

]2=√

2 (1 + i)

[−1 + i

√3

2

]2=

(√

3− 1)− (√

3 + 1)i√2

. �

11. Find the four roots of the equation z4 +4 = 0 and use them to factor z4 +4 into quadratic factorswith real coefficients.

Answer. The equation z4 + 4 = 0 is equivalent to z4 = −4, i.e., z = (−4)1/4. We express thecomplex numbers −4 and (−4)1/4 in the exponential form. Since −4 has the modulus | − 4| = 4and the principal argument Arg(−4) = π, because we use a principal argument in (−π, π]. Usingthe information, we have

−4 = | − 4|ei arg(−4) = 4ei(Arg(−4)+2kπ) = 4ei(π+2kπ) = 4e(2k+1)πi,

(−4)1/4 = 41/4e(2k+1)πi

4 =√

2e(2k+1)πi

4 , k = 0,±1,±2, . . . .

Hence, we deduce all four roots by putting k = 0, 1, 2, 3,

c0 =√

2eπi4 =√

2

(1√2

+1√2i

)= 1 + i, c1 =

√2e

3πi4 =

√2

(− 1√

2+

1√2i

)= −1 + i,

c2 =√

2e5πi4 =

√2

(− 1√

2− 1√

2i

)= −1− i, c3 =

√2e

7πi4 =

√2

(1√2− 1√

2i

)= 1− i.

Since c0, c1, c2, and c3 are roots of the equation z4 + 4 = 0, so we can express z4 + 4 as follows:

z4 + 4 = [z − c0] [z − c1] [z − c2] [z − c3]

= [z − (1 + i)] [z − (−1 + i)] [z − (−1− i)] [z − (1− i)]

= [(z − 1)− i] [(z + 1)− i] [(z + 1) + i] [(z − 1) + i]

=[(z − 1)2 − i2

][(z + 1)− i] [(z + 1) + i]

=[(z − 1)2 − i2

] [(z + 1)2 − i2

]=[z2 − 2z + 2

] [z2 + 2z + 2

]. �

12. Show that if c is any nth root of unity other than unity itself, then

1 + c+ c2 + · · ·+ cn−1 = 0.

Page 8 of 9


Proof. We observe that if zw = 0 for complex numbers z 6= 0 and w, then w = 0. For this reason,

(1− c)(1 + c+ c2 + · · ·+ cn−1

)= 0 implies 1 + c+ c2 + · · ·+ cn−1 = 0,

where c 6= 1. Hence, it is enough to prove that

(1− c)(1 + c+ c2 + · · ·+ cn−1

)= 0.

Expanding the left–hand side, we get

(1− c)(1 + c+ c2 + · · ·+ cn−1

)= 1 + c+ c2 + · · ·+ cn−1 − c

(1 + c+ c2 + · · ·+ cn−1

)= 1 + c+ c2 + · · ·+ cn−1 −

(c+ c2 + c3 + · · ·+ cn−1 + cn

)= 1 + c+ c2 + · · ·+ cn−1 −

(c+ c2 + c3 + · · ·+ cn−1 + 1

)= 0,

where cn = 1 is used. �

Another Proof. Using the identity in the problem 7 above, we have

1 + c+ c2 + · · ·+ cn−1 =1− cn

1− c=

1− 1

1− c= 0,

because c is the nth root of unity, i.e., cn = 1. �

Page 9 of 9




Complex Analysis IMATH 315 SECTION 01 CRN 235169:30 – 10:45 on Monday & WednesdayDue Date: Wednesday, October 7, 2009


1. Sketch the following sets and determine which are domains.

(1.1) | z − 2 + i | ≤ 1.

Answer. With z = x+ iy, the given inequality implies

1 ≤ | x− 2 + i ( 1 + y ) | = (x− 2 )2 + ( 1 + y )2 ,

i.e., it represents the region outside (including boundary) the circle centered at (2,−1) with the

radius 1. See the figure below. �(1.2) | 2z + 3 | > 4.


4 < | 2x+ 3 + 2yi | = ( 2x+ 3 )2 + ( 2y )2 , 1 <

(x+

3

2

)2

+ y2,

i.e., it represents the region outside (excluding boundary) the circle centered at (−3/2, 0) with

the radius 1. See the figure below. �

1.0 1.5 2.0 2.5 3.0

-2.0

-1.5

-1.0

-0.5

0.0

X

Y

H1.1L Èz-2+iÈ£1

-4 -3 -2 -1 0 1

-2

-1

0

1

2

X

Y

H1.2L È2z+3È>4

(1.3) Im(z) > 1.


1 < Im(z) = x

i.e., it represents the region on the left–hand side (excluding boundary) the line x = 1. See the

figure below. �(1.4) | z − 4 | ≥ | z |.


x2 + y2 = |x+ iy | ≤ |x− 4 + iy | = (x− 4 )2 + y2 = x2 − 8x+ 16 + y2, 8x ≤ 16, x ≤ 2,

i.e., it represents the region on the left–hand side (including boundary) the line x = 2. See the

figure below. �

Page 1 of 5


-3 -2 -1 0 1 2 3

0

1

2

3

4

5

6

X

Y

H1.3L ImHzL>1

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

X

Y

H1.4L Èz-4È³ÈzÈ

2. In each case, sketch the closure of the set.

(2.1) |Re(z) | < | z |.

Answer. With z = x+ iy, we observe

|x | = |Re(z) | < | z | =√

x2 + y2, i.e., x2 = |x |2 < x2 + y2, 0 < y2, 0 < | y | ,

which represents the whole complex plane except the real axis (y = 0). Hence, the closure of the

set will be the whole complex plane. �(2.2) Re(z2) > 0.

Answer. With z = x+ iy, we have

z2 = x2 − y2 + 2xyi, Re(z2) = x2 − y2.

So the region, 0 < x2 − y2 = (x− y)(x+ y), is the region satisfying (1) x− y > 0 and x+ y > 0and (2) x − y < 0 and x + y < 0. Hence, the closure of the set will be the region including the

boundaries y = ±x. See the figure below. �

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

X

Y

H2.2L ReHz2L>0

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

X

Y

H2.2L ReHz2L³0

Page 2 of 5


3. For each of the functions below, describe the domain of definition that is understood.

(3.1) f(z) = Arg

(1

z

).

Answer. Since 1/z is defined at any z, except at z = 0, thus f(z) is defined everywhere except

at z = 0. �

(3.2) f(z) =z

z + z̄.

Answer. The denominator of f(z) vanishes at the point satisfying

z + z̄ = 0, z = −z̄, i.e., x+ iy = − (x− iy ) = −x+ iy, i.e., x = 0,

i.e., f(z) is defined everywhere except at the points on the imaginary axis. �

(3.3) f(z) =1

1− | z |2.

Answer. The denominator of f(z) vanishes at the point satisfying

1− | z |2 = 0, | z |2 = 1, i.e., x2 + y2 = 1,

i.e., f(z) is defined everywhere except at the points on the unit circle centered at the origin. �

4. Suppose that f(z) = x2 − y2 − 2y + i ( 2x− 2xy ), where z = x+ yi. Use the expressions,

x =z + z̄

2, and y =

z − z̄

2i,

to write f(z) in terms of z and simplify the result.

Answer. We recall that

z2 = (x+ iy )2 = x2 − y2 +2ixy, Re(z2

)= x2 − y2, Im

(z2

)= 2xy, z2 = Re

(z2

)− i Im

(z2

).

Using them with the given expressions above, we have

f(z) = x2 − y2 − 2y + i ( 2x− 2xy )

= Re(z2

)− 2

z − z̄

2i+ 2i

z + z̄

2− i Im

(z2

)= Re

(z2

)+ i ( z − z̄ ) + i ( z + z̄ )− i Im

(z2

)= Re

(z2

)− i Im

(z2

)+ 2iz = z2 + 2iz. �

Page 3 of 5


5. Write the function

f(z) = z +1

z, z ̸= 0,

in the form f(z) = u(r, θ) + iv(r, θ).

Answer. Using the exponential form z = reiθ, we have

f(z) = reiθ +1

reiθ= reiθ +

e−iθ

r

= r ( cos θ + i sin θ ) +1

r( cos θ − i sin θ )

=

(r cos θ +

1

rcos θ

)+ i

(r sin θ − 1

rsin θ

)=

(r +

1

r

)cos θ + i

(r − 1

r

)sin θ.

Hence, in the form f(z) = u(r, θ) + iv(r, θ), we have

u(r, θ) =

(r +

1

r

)cos θ, v(r, θ) =

(r − 1

r

)sin θ. �

6. Find and sketch the image S ′ of the semi–infinite strip S = { z = (x, y) | 0 ≤ x, 0 ≤ y ≤ π } underthe given transformation.

(6.1) w = z2.

Answer. A simple computation shows with z = x+ iy,

u(x, y) + iv(x, y) = w = z2 = x2 − y2 + 2ixy, u(x, y) = x2 − y2, v(x, y) = 2xy.

(1) The line y = 0 corresponds to u(x, 0) = x2 and v(x, 0) = 0, and (u, v) = (x2, 0) makes thepositive real axis in the w–plane. That is, y = 0 corresponds to the positive real axis in thew–plane. As x decreases on the line y = 0, u = x2 also decreases on the positive real axis.

(2) The line x = 0 corresponds to u(0, y) = −y2 and v(0, y) = 0, and (u, v) = (−y2, 0) makesthe negative real axis in the w–plane. That is, x = 0 corresponds to the negative real axis in thew–plane. As y increases on the line x = 0, u = −y2 decreases on the negative real axis.

(3) The line y = π corresponds to u(x, π) = x2−π2 and v(x, π) = 2πx and (u, v) = (x2−π2, 2πx)makes a parabolic curve of equation

u =( v

2π

)2

− π2, 4π2(u+ π2

)= v2, v = ±2π

√u+ π2.

As x increases on the line y = π, both u = x2 − π2 and v = 2πx increases on the parabolic curvev = ±2π

√u+ π2.

(4) Since 0 ≤ x and 0 ≤ y ≤ π, so v = 2xy ≥ 0.

Combining (1) through (4), the image S ′ should be bounded by the real axis and the paraboliccurve, i.e., we have the image S ′ as follows:

S ′ ={w = (u, v) | 0 ≤ v ≤ 2π

√u+ π2

}. �

Page 4 of 5


-1 0 1 2 3

-1

0

1

2

3

4

x

y

H6.1L S=8Hx,yLÈ 0£x, 0£y£Π <

-15 -10 -5 0 5 10

-10

0

10

20

30

u

v

H6.1L S'=8Hu,vLÈ 0£v£2Π u + Π2 <

(6.2) w = ez.

Answer. A simple computation shows with z = x+ iy,

ρeiϕ = w = ez = ex+iy = exeiy, ρ = ex, ϕ = y.

(1) The line y = 0 corresponds to ϕ = 0 and ρ = ex, and so in polar coordinates system, itcorresponds to the positive real axis in the w–plane. As x decreases on the line y = 0, ρ = ex

also decreases on the positive real axis.

(2) The line x = 0 corresponds to ρ = e0 = 1 and ϕ = y, and so in polar coordinates system, itcorresponds to the circle with radius 1. As y increases on the line x = 0, ϕ = y increases on thecircle, i.e., the points on the circles moves counterclockwise.

(3) The line y = π corresponds to ϕ = π and ρ = ex, and so in polar coordinates system, itcorresponds to the negative real axis. As x increases on the line y = π, the point on the negativereal axis ρ = ex goes away from the origin, i.e., decreases.

Combining (1) through (3), the image S ′ should be above the real axis and outside the unit circlecentered at the origin, i.e., we have the image S ′ as follows:

S ′ ={w = ρeiϕ | 1 ≤ ρ, 0 ≤ ϕ ≤ π

}. �

-1 0 1 2 3

-1

0

1

2

3

4

x

y

H6.1L S=8Hx,yLÈ 0£x, 0£y£Π <

-2 -1 0 1 2

-1

0

1

2

3

u

v

H6.2L S'=8ΡãiΦÈ 1£Ρ, 0£Φ£Π <

Page 5 of 5




• Section 15 Limits

• Section 16 Theorem on Limits

• Section 17 Limits Involving The Point At Infinity

• Section 18 Continuity

Complex Analysis IMATH 315 SECTION 01 CRN 235169:30 – 10:45 on Monday & WednesdayDue Date: Monday, October 19, 2009


1. Compute the given complex limit.

(1.1) limz→ 2i

(z2 − z̄

)Answer. Since z2 and z̄ are continuous everywhere in the complex plane, so we have

limz→ 2i

(z2 − z̄

)= ( 2i )2 − 2i = −4− (−2i ) = −4 + 2i. �

(1.2) limz→ 1+i

z − z̄z + z̄

Answer. Since z − z̄ and z + z̄ are continuous everywhere in the complex plane and z + z̄ 6= 0at z = 1 + i, so we have

limz→ 1+i

z − z̄z + z̄

=limz→ 1+i ( z − z̄ )

limz→ 1+i ( z + z̄ )=

1 + i− ( 1− i )

1 + i+ 1− i=

2i

2= i. �

(1.3) limz→πi

ez

Answer. Since ez is continuous everywhere in the complex plane, so we have

limz→πi

ez = elimz→πi z = eπi = cosπ + i sin π = −1. �

(1.4) limz→ 2+i

( ez + z )

Answer. Since ez + z is continuous everywhere in the complex plane, so we have

limz→ 2+i

( ez + z ) = e2+i + 2 + i = e2ei + 2 + i

= e2 ( cos 1 + i sin 1 ) + 2 + i =(

2 + e2 cos 1)

+ i(

1 + e2 sin 1). �

(1.5) limz→ 1+i

z2 + 1

z2 − 1

Answer. Since z2 + 1 and z2− 1 are continuous everywhere in the complex plane and z2− 1 6= 0at z = 1 + i, so we have

limz→ 1+i

z2 + 1

z2 − 1=

limz→ 1+i ( z2 + 1 )

limz→ 1+i ( z2 − 1 )=

( 1 + i )2 + 1

( 1 + i )2 − 1=

2i+ 1

2i− 1=

3 + 4i

5. �

(1.6) limz→−i

z4 − 1

z + i

Answer. We observe

z4 − 1

z + i=

(z2 + 1)(z2 − 1)

z + i=

(z + i)(z − i)(z2 − 1)

z + i= (z − i)(z2 − 1),

which is a polynomial. So we have

limz→−i

z4 − 1

z + i= lim

z→−i(z − i)(z2 − 1) = (−2i )

[(−i )2 − 1

]= 4i. �

Page 1 of 5


2. Consider the limit limz→ 0

Re(z)

Im(z).

(2.1) What value does the limit approach as z approaches 0 along the line y = x?

Answer. As z = (x, y) moves along the line y = x, we have Re(z) = x = Im(z), which implies

limz→ 0

Re(z)

Im(z)= lim

x→ 0

x

x= lim

x→ 01 = 1. �

(2.2) What value does the limit approach as z approaches 0 along the imaginary axis?

Answer. As z = (x, y) moves along the imaginary axis, i.e., the line x = 0, we have Re(z) = 0and Im(z) = y, which implies

limz→ 0

Re(z)

Im(z)= lim

y→ 0

0

y= lim

y→ 00 = 0. �

(2.3) Based on your answers for (2.1) and (2.2), what can you say about limz→ 0

Re(z)

Im(z)?

Answer. By the uniqueness of the limit, the answers for (2.1) and (2.2) imply that the limit

does not exist. �

3. Consider the limit limz→ i

( | z |+ iArg ( iz ) ).

(3.1) What value does the limit approach as z approaches i along the unit circle | z | = 1 in the firstquadrant?

Answer. For z = eiθ on the unit circle 〈 z 〉 = 1, we observe

iz = eiπ2 eiθ = ei(

π2+θ ).

So when z is in the first quadrant, i.e., 0 ≤ θ ≤ π

2, we have

π

2≤ π

2+ θ ≤ π,

i.e., iz is in the second quadrant and by the same argument, when z is in the second quadrant,iz is in the third quadrant.

Now, as z approaches i along the unit circle | z | = 1 in the first quadrant, iz approaches −1 inthe second quadrant. So we have

limz→ i

( | z |+ iArg ( iz ) ) = | i |+ iπ = 1 + iπ. �

(3.2) What value does the limit approach as z approaches i along the unit circle | z | = 1 in the secondquadrant?

Answer. As z approaches i along the unit circle | z | = 1 in the second quadrant, iz approaches−1 in the third quadrant. So we have

limz→ i

( | z |+ iArg ( iz ) ) = | i |+ i (−π ) = 1− iπ. �

(3.3) Based on your answers for (3.1) and (3.2), what can you say about limz→ i

( | z |+ iArg ( iz ) )?

Answer. By the uniqueness of the limit, the answers for (3.1) and (3.2) imply imply that the

limit does not exist. �

Page 2 of 5


4. Compute the given limit.

(4.1) limz→∞

z2 + iz − 2

(1 + 2i)z2

Answer. Letting f(z) =z2 + iz − 2

(1 + 2i)z2, we compute f(1/z):

f

(1

z

)=

(1/z)2 + i/z − 2

(1 + 2i)/z2=

1 + iz − 2z2

1 + 2i,

limz→ 0

f

(1

z

)= lim

z→ 0

1 + iz − 2z2

1 + 2i=

1

1 + 2i=

1− 2i

5.

Hence, by the Theorem, we deduce

limz→∞

f(z) = limz→∞

z2 + iz − 2

(1 + 2i)z2=

1− 2i

5. �

(4.2) limz→ i

z2 − 1

z2 + 1

Answer. Letting f(z) =z2 − 1

z2 + 1, we have

1

f(z)=z2 + 1

z2 − 1, lim

z→ i

1

f(z)= lim

z→ i

z2 + 1

z2 − 1=

0

−2= 0.


limz→ i

f(z) = limz→ i

z2 − 1

z2 + 1=∞. �

(4.3) limz→∞

z2 − (2 + 3i)z + 1

iz − 3

Answer. Letting f(z) =z2 − (2 + 3i)z + 1

iz − 3, we have

f

(1

z

)=

(1/z)2 − (2 + 3i)/z + 1

i/z − 3=

1− (2 + 3i)z + z2

iz − 3z2,

1

f(1/z)=

iz − 3z2

1− (2 + 3i)z + z2,

limz→ 0

1

f(1/z)= lim

z→ 0

iz − 3z2

1− (2 + 3i)z + z2=

0

1= 0.


limz→∞

f(z) = limz→∞

z2 − (2 + 3i)z + 1

iz − 3=∞. �

Page 3 of 5


5. Show that the function f is continuous at the given point.

(5.1) f(z) = z3 − 1

z; z0 = 3i

Proof. With z = x+ iy, we observe

f(x+ iy) = (x+ iy )3 − 1

x+ iy

= x3 − 3xy2 − x

x2 + y2+ i

(3x2y − y3 +

y

x2 + y2

)= u(x, y) + iv(x, y).

Since u(x, y) and v(x, y) are continuous at (x, y) = (0, 3) from Calculus/Real Analysis, hence, by

Theorem, we conclude f(z) is continuous at z0 = 3i. One may prove in a different way. �

(5.2) f(z) =

z3 − 1

z − 1, | z | 6= 1

3, | z | = 1; z0 = 1

Proof. We compute the limit

limz→ z0

f(z) = limz→ 1

z3 − 1

z − 1= lim

z→ 1

(z − 1)(z2 + z + 1)

z − 1= lim

z→ 1

(z2 + z + 1

)= 3 = f(1) = f(z0).

Hence, by the definition, f(z) is continuous at z0 = 1. �

6. Show that the function f is discontinuous at the given point.

(6.1) f(z) = Arg ( iz ); z0 = i

Proof. (1) As z approaches z0 = i along the unit circle | z | = 1 in the first quadrant, iz approaches−1 along the unit circle | z | = 1 in the second quadrant and so Arg ( iz ) moves to π.

(2) As z approaches z0 = i along the unit circle | z | = 1 in the second quadrant, iz approaches−1 along the unit circle | z | = 1 in the third quadrant and so Arg ( iz ) moves to −π.

From (1) and (2), the limit is not unique and thus the function f is not continuous at z0 = i. �

(6.2) f(z) =

z3 − 1

z − 1, | z | 6= 1

3, | z | = 1; z0 = i

Proof. We compute the limit

limz→ z0

f(z) = limz→ i

z3 − 1

z − 1= lim

z→ i

(z − 1)(z2 + z + 1)

z − 1= lim

z→ i

(z2 + z + 1

)= i 6= 3 = f(i) = f(z0).

Hence, by the definition, f(z) is not continuous at z0 = i. �

Page 4 of 5


7.(7.1) Is it true that limz→ z0

f(z) = limz→ z0

f ( z̄ ) for any complex function f? If so, give a brief justification.

If not, find a counterexample.

Answer. No. We recall from the Example 2 on page 47 in the textbook that f(z) =z

z̄is not

continuous at z = 0, because the limit does not exist there. With this function and z0 = 0, weobserve

f(z) =( zz̄

)=z̄¯̄z

=z̄

z= f ( z̄ ) .

However, since the limit limz→ 0

z̄

zdoes not exist (by the similar argument as in the Example 2 on

page 47), so we cannot say limz→ 0

f(z) = limz→ 0

f ( z̄ ). �

(7.2) If f(z) is a continuous function at z0, then is it true that f(z) is continuous at z0?

Answer. Yes. We recall the theorem saying that a composition of two continuous functions is alsocontinuous. Since g(z) = z̄ is continuous everywhere, for a function f which is continuous at z0, the

composition ( g ◦ f ) (z) = g(f(z)) = f(z) should be continuous at z0. �

8. If f satisfies limx→ 0

f(x+ i0) = 0 and limy→ 0

f(0 + iy) = 0, then can you conclude that limz→ 0

f(z) = 0? Ex-

plain.

Answer. No. Consider the function f(z) =( zz̄

)2− 1. We observe

f(x+ iy) =

(x+ iy

x− iy

)2

− 1,

f(x+ i0) =( xx

)2− 1 = 0, f(0 + iy) =

(iy

−iy

)2

− 1 = (−1 )2 − 1 = 0,

which implieslimx→ 0

f(x+ i0) = limx→ 0

0 = 0, limy→ 0

f(0 + iy) = limy→ 0

0 = 0.

However, as z = (x, y) approaches 0 along the line y = x, we observe

limz→ 0

f(z) = lim(x,x)→ (0,0)

f(x+ ix) = lim(x,x)→ (0,0)

[(x+ ix

x− ix

)2

− 1

]

= lim(x,x)→ (0,0)

[(1 + i

1− i

)2

− 1

]=

(1 + i

1− i

)2

− 1 = −2 6= 0. �

Page 5 of 5

United Arab Emirates University

College of Sciences


HOMEWORK 5 { SOLUTION

� Section 19 Derivatives

� Section 20 Di�erentiation Formulas

Complex Analysis I

MATH 315 SECTION 01 CRN 23516

9:30 { 10:45 on Monday & Wednesday

Due Date: Monday, November 2, 2009

ID No: Solution

Name: Solution

Score: Solution


1. (#1 on Page 62) Use results in Sec. 20 to �nd f 0(z) when

(1.1) (b) f(z) =�1� 4z2

�3

Answer.

f 0(z) = 3�1� 4z2

�2�1� 4z2

�0

= �24z�1� 4z2

�2

: �

(1.2) (c) f(z) =z � 1

2z + 1, z 6= �1=2

Answer.

f 0(z) =2z + 1� (z � 1) 2

(2z + 1)2=

3

(2z + 1)2: �

Page 1 of 3


2. (#8 on Page 63) Use the method in Example 2, Sec. 19, to show that f 0(z) does not exist at any

point z when

(2.1) (a) f(z) = Re(z)

Proof.4w

4z=

Re (z +4z)� Re (z)

4z=

x+4x� x

4x+ i4y=

4x

4x+ i4y;

where 4z = 4x+ i4y and z = x+ iy.

As 4z approaches (0; 0) horizontally through the points (4x; 0) on the real axis, we have

4w

4z=

4x

4x+ i4y=

4x

4x+ i0= 1:

As 4z approaches (0; 0) horizontally through the points (0;4y) on the imaginary axis, we have

4w

4z=

4x

4x+ i4y=

0

0 + i4y= 0:

Since the limit is not unique, thus, f 0(z) does not exist at any point. �

(2.2) (b) f(z) = Im(z)

Proof.4w

4z=

Im (z +4z)� Im (z)

4z=

y +4y � y

4x+ i4y=

4y

4x+ i4y;

where 4z = 4x+ i4y and z = x+ iy.

As 4z approaches (0; 0) horizontally through the points (4x; 0) on the real axis, we have

4w

4z=

4y

4x+ i4y=

0

4x+ i0= 0:

As 4z approaches (0; 0) horizontally through the points (0;4y) on the imaginary axis, we have

4w

4z=

4y

4x+ i4y=

4y

0 + i4y=

1

i:

Since the limit is not unique, thus, f 0(z) does not exist at any point. �

Page 2 of 3


3. (#9 on Page 63) Let f denote the function whose values are

f(z) =

8><>:

�z2

zwhen z 6= 0,

0 when z = 0.

(1) Show that if z = 0, then4w

4z= 1 at each nonzero point on the real and imaginary axes in the

4z, or 4x4y, plane.

Proof. Case 1. z = 0:

4w

4z=

f(z +4z)� f(z)

4z=

z +4z2

= (z +4z)� 0

4z=

z +4z2

4z (z +4z)

where 4z = 4x+ i4y and z = x+ iy. �

(2) Show that4w

4z= �1 at each nonzero point (4x;4x) on the line 4y = 4x in that plane.

(3) Conclude form these observations that f 0(0) does not exist. Note that to obtain this result, it

is not su�cient to consider only horizontal and vertical approaches to the origin in the 4z plane.

(Compare with Example 2, Sec. 19.)

Page 3 of 3

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