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Solutions to Selected Exercises

in

Complex Analysis with

Applications

by N. Asmar and L. Grafakos

Section 1.1 Complex Numbers 1

Solutions to Exercises 1.

1. We have

1 − i 2 =^

2 + (−^

2 )i.

So a =

2 and^ b^ =^ −^

5. We have

(2 − i)^2 = (2 + i)^2 (because 2 − i = 2 − (−i) = 2 + i)

= 4 + 4i +

︷︸︸︷^ =−^1

(i)^2 = 3 + 4i.

So a = 3 and b = 4.

9. We have

2 +^

i 7

2 −^ i

2 +^ i

14 −^

= (^17) ︷ ︸︸ ︷ −i

i 7 =

28 −^ i^

So a = 2528 and b = − 27.

13. Multiplying and dividing by the conjugate of the denominator, i.e. by 2 − i = 2 + i we get

14 + 13i 2 − i

= (14 + 13i)(2 +^ i) (2 − i)(2 + i)

=^14 ·^ 2 + 14^ ·^ i^ + 13i^ ·^ 2 + 13i

2 4 + 1 =

28 + 14i + 26i − 13 5 =

5 i^ = 3 + 8i

So a = 3 and b = 8.

17. Multiplying and dividing by the conjugate of the denominator, i.e. by x − iy = x + iy we get

x + iy x − iy =^

x + iy x − iy ·^

(x + iy) (x + iy)

= (x^ +^ iy)

2 x^2 + y^2 = x

(^2) + 2xyi + y (^2) i 2 x^2 + y^2 = x

(^2) − y (^2) + 2xyi x^2 + y^2 = x

(^2) − y 2 x^2 + y^2

• 2 xy x^2 + y^2

i.

Section 1.1 Complex Numbers 3

(2 + 3i)z = (2 − i)z − i {(2 + 3i) − (2 − i)}z = −i (2 + 3i − 2 + i)z = −i 4 iz = −i

z = −

i 4 i z = −

So z = − 1 4

iz + 2i = 4 iz + 2i = 4 (Conjugating both sides) iz + 2i = 4 (Using problem 34) iz = 4 − 2 i

z =

4 − 2 i i z =

4 − 2 i i ·

−i −i

z =

− 4 i + 2i^2 1 z = − 2 − 4 i

So z = − 2 − 4 i.

33. We are given

(1 − i)z 1 + z 2 = 3 + 2i z 1 + (2 − i)z 2 = 2 + i

From the first equation we obtain

z 2 = 3 + 2i − (1 − i)z 1 = 3 + 2i − z 1 + iz 1

4 Chapter 1 A Preview of Applications and Techniques

We then substitute the expression for z 2 into the second equation to get

z 1 + (2 − i)(3 + 2i − z 1 + iz 1 ) = 2 + i z 1 + 6 + 4i − 2 z 1 + 2iz 1 − 3 i − 2 i^2 + iz 1 − i^2 z 1 = 2 + i z 1 + 6 + 4i − 2 z 1 + 2iz 1 − 3 i + 2 + iz 1 + z 1 = 2 + i 3 iz 1 + 8 + i = 2 + i 3 iz 1 = 2 + i − 8 − i 3 iz 1 = − 6 z 1 = −^6 3 i z 1 = − 2 i

Thus

z 2 = 3 + 2i − 2 i + 2i^2 = 3 + 2i − 2 i − 2 = 1

So z 1 = 2i and z 2 = 1.

37. We have

x^2 + 4x + 5 = 0

x = −^4 ±

42 − 4 × 5

(By the quadratic formula)

=

− 4 ± 2 i 2 = − 2 ± i

41. We have

anzn^ + an− 1 zn−^1 + · · · + a 1 z + a 0 = anzn^ + an− 1 zn−^1 + · · · + a 1 z + a 0 = an · zn^ + an− 1 · zn−^1 + · · · + a 1 · z + a 0 [Since a 0 , a 1 , · · · , an− 1 , an are all real] = anzn^ + an− 1 zn−^1 + · · · + a 1 z + a 0 [by using the property that zn^ = (z)n] = an(z)n^ + an− 1 (z)n−^1 + · · · + a 1 (z) + a 0.

45. We already know that z = 1 + i is a root of p(z) = z^4 + 4, then then by problem 48, z = 1 − i is also a root of p(z). Thus h(z) = (z − 1 − i)(z − 1 + i) = z^2 − 2 z + 2 divides p(z)

6 Chapter 1 Complex Numbers and Functions

Solutions to Exercises 1.

1. Note that z = 1−i, −z = −(1−i) = −1+i and ¯z = 1 − i = 1+i. Thus z has coordinates (1, −1), −z has coordinates (− 1 , 1) and ¯z has coordinates (1, 1). Also |z| =

12 + 1^2 =

5. Note that z = 1 − i = 1 + i. Hence −z = −(1 + i) = − 1 − i and z = 1 + i = 1 − i. Thus z has coordinates (1, 1), −z has coordinates (− 1 , −1) and z has coordinates (1, −1). Also |z| =

12 + 1^2 =

9. We use the property that |ab| = |a||b| twice below:

|(1 + i)(1 − i)(1 + 3i)| = |1 + i||(1 − i)(1 + 3i)| =

√ 2 ︷ ︸︸ ︷ |1 + i|

√ 2 ︷ ︸︸ ︷ | 1 − i|

√ 10 ︷ ︸︸ ︷ |1 + 3i| = 2

13. Using the properties

∣ z z^12

∣ =^ ||zz^12 || and^ |z|^ =^ |z|^ we have ∣∣ ∣∣^ i 2 − i

∣∣ = |i| | 2 − i|

|i| | 2 − i| =^

√^1

Section 1.2 The Complex Plane 7

17. The equation |z − i| = −1 does not have any solution because |z| is a distance from the point z to the origin. And clearly a distance is always non-negative.
18. Let z = x + iy and we have

|z − 1 | ≤ 4 |x + iy − 1 | ≤ 4 |(x − 1) + iy|^2 ≤ 42 (Squaring both sides) (x − 1)^2 + y^2 ≤ 16

The inequality |z − 1 | ≤ 4 represents a closed disc with radius 4 units and centre at z = 1 as shown below.

25. Let z = x + iy and we have

0 < |z − 1 − i| < 1 0 < |x + iy − 1 − i| < 1 0 < |(x − 1) + i(y − 1)|^2 < 12 0 < (x − 1)^2 + (y − 1)^2 < 1

The inequality 0 < |z − 1 − i| < 1 represents a puntured open disc with radius 1 units, centre at z = 1 + i and puctured at z = 1 + i as shown below.

Section 1.2 The Complex Plane 9

From the second equation we have

y − y 1 = t(y 2 − y 1 ) y − y 1 y 2 − y 1 =^ t

Now by eliminating t from both the equations, we get

x − x 1 x 2 − x 1

= y^ −^ y^1 y 2 − y 1 x − x 1 y − y 1 =^

x 2 − x 1 y 2 − y 1

which represents the equation of a line passing through distinct points (x 1 , y 1 ) and (x 2 , y 2 ).

33. We know that

z.z = |z|^2

z.

z |z|^2

= 1 (Dividing both sides by |z|^2 )

Thus we get that z−^1 = z |z|^2

37. (a) We get the estimate

| cos θ + i sin θ| ≤ | cos θ| + |i sin θ|

by the triangle inequality. Now we notice that

|i sinθ| = |i| | sin θ| = 1| sin θ| = | sin θ|.

So, | cos θ| + |i sin θ| = | cos θ| + | sin θ|.

We know from algebra that | sin θ| ≤ 1, and | cos θ| ≤ 1 Therefore we have

| cos θ| + | sin θ| ≤ 1 + 1 = 2,

and this justifies the last step of the estimation above. (b) By definition of the absolute value we know that if z = x + iy then |z| =

x^2 + y^2. Therefore if z = cos θ + i sin θ then |z| =

(cos θ)^2 + (sin θ)^2. But from trigonometry we know the formula (cos θ)^2 + (sin θ)^2 = 1

is true for any number θ. Therefore we have

| cos θ + i sin θ| =

(cos θ)^2 + (sin θ)^2 = 1.

10 Chapter 1 Complex Numbers and Functions

41. We want to know something about | (^) z−^14 | given some information about |z − 1 |. Notice that an upper bound of the quantity | (^) z−^14 | is given by the reciprocal of any lower bound for |z − 4 |. So, first we can find some information about |z − 4 |. For this we use some ideas from the example 8 in this section. We notice that z − 4 = (z − 1) − 3. Therefore by the inequality |z 1 − z 2 | ≥ ||z 1 | − |z 2 || we have

|z − 4 | = |(z − 1) − 3 | ≥ ||z − 1 | − 3 |

Now since we know |z− 1 | ≤ 1 we get that ||z− 1 |− 3 | ≥ | 1 − 3 | = 2. Hence |z− 4 | ≥ 2. Finally taking reciprocals of the sides of the inequality we will reverse the sign of the inequality and get the needed upper estimate: 1 |z − 4 |

45. (a) By triangle inequality we have ∣∣ ∣∣ ∣∣

∑^ n

j=

vj wj

∑^ n

j=

|vj wj |.

And now we use the properties |z 1 z 2 | = |z 1 ||z 2 | and |z| = |z| and get

|vj wj | = |vj ||wj | = |vj ||wj |.

Using this identity in the triangle inequality above and recalling the assumption that we already proved () we have ∣∣ ∣∣ ∣∣

∑^ n

j=

vj wj

∑^ n

j=

|vj wj |

∑^ n

j=

|vj ||wj | ≤

√√∑^ n j=

|vj |^2

√√∑^ n j=

|wj |^2.

(b) Using the hint we start from the obvious inequality

∑^ n

j=

|vj | − |wj |

Expanding the right hand side of this inequality we have

∑^ n

j=

|vj | − |wj |

∑^ n

j=

|vj |^2 − 2 |vj ||wj | + |wj |^2

∑^ n

j=

|vj |^2 + |wj |^2

∑^ n

j=

|vj ||wj |.

12 Chapter 1 Complex Numbers and Functions

Solutions to Exercises 1.

1. We need to present the number given in its polar form in the form with the real and imaginary parts z = x + iy. We have

z = 3

cos^7 π 12

• i sin^7 π 12

= 3 cos^7 π 12

• i 3 sin^7 π 12

In Cartesian coordinates z is represented by (3 cos 712 π , 3 sin 712 π ) as shown below.

5. Let z = − 3 − 3 i. Then we have r =

(−3)^2 + (−3)^2 =

2. Also, we can find the argument by evaluating

cos θ = x r

= −^3

and sin θ = y r

= −^3

From the Table 1 in the Section 1.3 we see that θ = 54 π. Thus, arg z = 54 π + 2kπ. Since 54 π is not from the interval (−π, π] we can subtract 2π and get that Arg z = 54 π − 2 π = − 34 π. So, the polar representation is

− 3 − 3 i = 3

cos

3 π 4

• i sin

3 π 4

9. Again we can denote z = − 2 i. Then we have r =

(− 12 )^2 = 12. We can evaluate

cos θ = x r

= 0 and sin θ = y r

= −^1 /^2

And, from the Table 1 in the Section 1.3 we find θ = 32 π. (Also we could plot the complex number (− i 2 ) as a point in the complex plane and find the angle from the picture.) Therefore

Section 1.3 Polar form 13

arg z = 32 π + 2πk. Since 32 π is not from the interval (−π, π] we can subtract 2π and get that Arg z = 32 π − 2 π = − π 2. So, the polar representation is

− i 2

=^1

cos

− π 2

• i sin

− π 2

13. We have z = x + i y = 13 + i 2. Since x > 0 we compute

Arg z = tan−^1 (

y x ) = tan

− 1 (^2

13 )^ ≈^0.^153.

Hence we express arg z ≈ 0 .153 + 2πk for all integer k.

17. First, we need to express the number z = −

3 + i in the polar form. We compute r =

3 + 1 = 2. And since −

3 < 0, and 1 > 0 we see that Arg z = tan−^1

1 −√ 3

tan−^1

√ 3 3

= 56 π. Since we need to find the cube of the number z we can use the De Moivre’s Identity to get the polar representation:

3 + i)^3 =

cos

5 π 6

• i sin

5 π 6

cos

5 π 6

• i sin

5 π 6

cos

15 π 6

• i sin

15 π 6

cos

3 π 6

• i sin

3 π 6

Where in the last identity we used the fact that Arg (−

3 + i)^3 = 156 π − 2 π = 36 π should be in the interval (−π, π].

21. First, we find the modulus and the argument of the number z = 1 + i. We have r =

12 + 1^2 =

2. And, since for z = x + iy = 1 + i1 we get x > 0 we compute Arg z = tan−^1

( (^) y x

= tan−^1

1

= tan−^1 (1) = π 4. Hence z can be expressed in the polar form

1 + i =

cos

( (^) π 4

• i sin

( (^) π 4

After this we can use De Moivre’s identity to get

(1 + i)^30 =

cos

( (^) π 4

• i sin

( (^) π 4

2)^30

cos

30 π 4

• i sin

30 π 4

= 215 (0 + i · 1) = 2^15 i.

Thus Re

(1 + i)^30

= 0, and Im

(1 + i)^30

= 2^15.

Section 1.3 Polar form 15

37. In order to solve this equation we need to find the 7-th root of (−7). Thus we express (−7) in the polar form. −7 = 7(cos(π) + i sin(π)). Hence by the formula for the n-th root with n = 7 we have that the solutions are

z 1 = 7

7 (cos(π/7) + i sin(π/7)) , z 2 = 7

7 (cos(3π/7) + i sin(3π/7)) , z 3 = 7

7 (cos(5π/7) + i sin(5π/7)) , z 4 = 7

7 (cos(7π/7) + i sin(7π/7)) , z 5 = 7

7 (cos(9π/7) + i sin(9π/7)) , z 6 = 7

7 (cos(11π/7) + i sin(11π/7)) , and z 7 = 7

7 (cos(13π/7) + i sin(13π/7)).

16 Chapter 1 Complex Numbers and Functions

41. We have

z^4 = − 1 − i

z^4 =

cos^5 π 4

• i sin^5 π 4

So by the formula for the n-th roots with n = 4, we find the roots to be

z 1 = 8

cos^5 π 16

• i sin^5 π 16

z 2 = 8

cos

13 π 16 +^ i^ sin

13 π 16

z 3 = 8

cos

21 π 16 +^ i^ sin

21 π 16

z 4 = 8

cos^2916 π + i sin^2916 π

45. We have

(z + 2)^3 = 3i w^3 = 3

cos

π 2 +^ i^ sin^

π 2

(Set w = z + 2)

18 Chapter 1 Complex Numbers and Functions

So the principal square root is

w =

cos θ 2 + i sin θ 2

Now we use the half-angle identity i.e. cos θ 2

1 + cos θ 2

, sin θ 2

1 − cos θ 2

and compute

w =

2 +^ i

√√ 1 +^3

= 1 + 2i.

Thus the original solutions are

z 1 =

−1 + w 2 =^ i z 2 =

− 1 − w 2 =^ −^1 −^ i.

53. We have

z^4 − (1 + i)z^2 + i = 0 u^2 − (1 + i)u + i = 0 (Set u = z^2 )

u =

−(1 + i) ±

(1 + i)^2 − 4 i 2 (Using the quadratic formula) u = −(1 +^ i)^ ±

− 2 i 2 u = −(1 +^ i)^ ±^ (−1 +^ i) 2

(The principal square root of − 2 i is −1 + i, see problem 51) u = − 1 , −i.

Section 1.3 Polar form 19

Since z =

u, we get

z 1 =

= i z 2 = −

= −i z 3 =

−i

=

cos^32 π + i sin^32 π

= cos^34 π + i sin^34 π

= −

√^1

• i

√^1

z 4 = −

−i = −z 3 = √^1 2

− i √^1 2

57. By De Moivre’s identity for n = 3 we have

cos 3θ + i sin 3θ = (cos θ + i sin θ)^3 = cos^3 θ + 3i cos^2 θ sin θ + 3i^2 cos θ sin^2 θ + i^3 sin^3 θ = (cos^3 θ − 3 cos θ sin^2 θ) + i(3 cos^2 θ sin θ − sin^3 θ)

Now by comparing the real and imaginary parts we get

cos 3θ = cos^3 θ − 3 cos θ sin^2 θ sin 3θ = 3 cos^2 θ sin θ − sin^3 θ.

61. We have zn^ = 1 and by the formula for n-th roots we get

zn^ = cos 0 + i sin 0

ωk = cos

2 kπ n

• i sin

2 kπ n

(where k = 0, · · · , n − 1).

65. We prove the binomial formula for complex numbers by induction on n. The statement for n = 1 can be easily verified. By induction hypothesis, we assume that the statement is true for n and we prove it for n + 1.

(a + b)n+1^ = (a + b)n.(a + b)

=

( (^) n ∑

m=

n m

an−m.bm

(a + b)