MCS 351 ENGINEERING MATHEMATICS SOLUTION OF ...

MCS 351 ENGINEERING MATHEMATICS
SOLUTION OF CHAPTER 13-14-15
Prof. Dr. İsmet KARACA
ERWIN KREYSZIG, 9TH EDITION
NOVEMBER 2012

CHAPTER 13 - COMPLEX NUMBERS AND FUNCTIONS
13.1
1. We can write −i as 0 + i(−1). Then,
• i 2 = [0 + i(−1)][0 + i(−1)] = 0 − 1 + i.(0 + 0) = −1,
• i 3 = i 2 .i = (−1).i = −i,
• i 4 = i 3 .i = (−i).i = −i 2 = −(−1) = 1,
• i 5 = i 4 .i = 1.i = i
We know that i = −i.
• 1 i = −i
−i 2
= −i
1 = −i,
• 1 i 2 = 1
−1 = −1,
• 1 i 3 = 1 −i = i
−i 2 = i 1 = i.
2. Let z = 2 + 2i. Then we obtain iz = i(2 + 2i) = −2 + 2i.
Let z = −1 − 5i. Then iz = i(−1 − 5i) = 5 − i.
Let z = 4 − 3i. Then iz = i(4 − 3i) = 3 + 4i.
a + b = 90 ◦
1

a + b = 90 ◦
3.
z 1
= x 1 + iy 1
= (x 1 + iy 1 )(x 2 − iy 2 )
z 2 x 2 + iy 2 (x 2 + iy 2 )(x 2 − iy 2 )
= x 1x 2 − x 1 iy 2 + iy 1 x 2 − iy 1 iy 2
x 2 2 − ix 2y 2 + ix 2 y 2 − i 2 y 2 2
= x 1x 2 + y 1 y 2 + i(x 1 y 2 + x 2 y 1 )
x 2 2 + y2 2
= x 1x 2 + y 1 y 2
x 2 2 + y2 2
+ i. x 2y 1 + x 1 y 2
x 2 2 + y2 2
4. We assume that z 1 ≠ 0.
0 = 0 z 1
= z 1z 2
z 1
= z 2
That is, if z 1 ≠ 0 then z 2 = 0.
Similarly you can show that if z 2 ≠ 0 then z 1 must be zero. So, at least one factor must be zero.
5. (⇒) If z = x + iy is pure imaginary then x = 0. So z = iy and z = −iy. That is z = −z.
(⇐) Let z = −z.
z = x − iy = −(x + iy) = −x − iy
x must be zero because x ∈ R and x = −x. So z is pure imaginary.
6. z 1 = 24 + 10i, z 2 = 4 + 6i
• z 1 +z 2 = (24 + 10i)+(4 + 6i) = (24−10i)+(4−6i) = 28−16i = (28 + 16i) = [(24 + 10i) + (4 + 6i)]
= (z 1 + z 2 )
• z 1 −z 2 = (24 + 10i)−(4 + 6i) = (24−10i)−(4−6i) = 20−4i = (20 + 4i) = [(24 + 10i) − (4 + 6i)]
= (z 1 − z 2 )
• z 1 .z 2 = (24 + 10i).(4 + 6i) = (24−10i).(4−6i) = 36−184i = (36 + 184i) = [(24 + 10i).(4 + 6i)]
= (z 1 .z 2 )
• z 1
z 2
= ( 24+10i
4+6i ) = ( (24+10i)(4−6i)
16+36
) = ( 156
52 − 104
52
156
i) =
52 + 104
52 i = (24+10i) = z 1
(4+6i) z 2
2

7-15: Complex Aritmetic Let z 1 = 2 + 3i and z 2 = 4 − 5i.
7. (5z 1 + 3z 2 ) 2 = [5.(2 + 3i) + 3.(4 − 5i)] 2 = [(10 + 15i) + (12 − 15i)] 2 = (22) 2 = 484
8. z 1 .z 2 = (2 − 3i).(4 + 5i) = 23 − 2i
9. Re( 1
1
−5−12i
) = Re(
z1
2 −5+12i
) = Re(
169
) = Re(− 5
169 − 12
169 i) = − 5
169
10.• z2 2 = (4 − 5i)2 = −9 − 40i ⇒ Re(z2 2) = −9
11. z 2
z 1
• (Rez 2 ) 2 = 4 2 = 16
12. • z 1
z 2
= 4−5i
2+3i = (4−5i)(2−3i)
4+9
= −7−22i
13
= − 7
13 − 22
13 i
= (2+3i)
(4−5i) = 2−3i
4+5i = −7−22i
41
= − 7
41 − 22
41 i
• ( z 1
z 2
) = ( 2+3i
4−5i ) = ( −7+22i
41
) = (− 7
41 + 22
41 i) = − 7
41 − 22
41 i
13. (4z 1 − z 2 ) 2 = [4.(2 + 3i) − (4 − 5i)] 2 = (8 + 12i − 4 + 5i) 2 = (4 + 17i) 2 = −273 + 136i
14. • z 1
z 1
= 2−3i
2+3i = −5−12i
4+9
= − 5
13 − 12
13 i
• z 1
z 1
= 2+3i
2−3i = −5+12i
4+9
= − 5
13 + 12
13 i
15. z 1+z 2
z 1 −z 2
= (2+3i)+(4−5i)
(2+3i)−(4−5i) = 6−2i
−2+8i = −28−44i
68
= − 7
17 − 11
17 i
16-19: Let z = x + iy.
16. • Im(z 3 ) = Im((x + iy) 3 ) = Im((x 3 − 3xy 2 ) + i.(3x 2 y − y 3 )) = 3x 2 y − y 3
• (Imz) 3 = (Im(x + iy)) 3 = y 3
17. 1 z = 1
x−iy = x+iy = x y
+ i ⇒ Re( 1 x 2 +y 2 x 2 +y 2 x 2 +y 2 z ) = x
18. At first we will find (1 + i) 8 :
(1 + i) 2 = 2i,
(1 + i) 3 = (1 + i) 2 .(1 + i) = 2i.(1 + i) = −2 + 2i,
(1 + i) 4 = (1 + i) 3 .(1 + i) = (−2 + 2i).(1 + i) = −4,
(1 + i) 5 = (1 + i) 4 .(1 + i) = (−4).(1 + i) = −4 − 4i,
(1 + i) 6 = (1 + i) 5 .(1 + i) = (−4 − 4i).(1 + i) = −8i,
(1 + i) 7 = (1 + i) 6 .(1 + i) = (−8i).(1 + i) = 8 − 8i,
(1 + i) 8 = (1 + i) 7 .(1 + i) = (8 − 8i).(1 + i) = 16.
x 2 +y 2
Then, Im((1 + i) 8 .z 2 ) = Im(16.z 2 ) = Im(16.(x 2 − y 2 + i2xy)) = Im(16(x 2 − y 2 ) + i32xy) = 32xy
19. Re( 1
z 2 ) = Re(
1
(x−iy) 2 ) = Re(
1
x 2 −y 2 −i2xy ) = Re( x2 −y 2 +i2xy
x 4 +2x 2 y 2 +y 4 ) =
x 2 −y 2
x 4 +2x 2 y 2 +y 4 .
13.2
1-8 Represent the followings in the polar form:
1. Let z = 3 − 3i. Then |z| = √ 3 2 + (−3) 2 = √ 9 + 9 = √ 18 = 3 √ 2 and since tan θ = 3
−3 = −1,
we get
⇒ θ = 7π 4 ⇒ z = 3√ 2(cos( 7π 4 ) + i. sin(7π 4 )).
3

the fact that tan θ = π 2 , θ = arctan( π 2 )
2. Let z = 2i. We can say that |z| = √ 2 2 = √ 4 = 2. Because of θ = π 2
, the result is
z = 2(cos( π 2 ) + i. sin(π 2 )).
On the other hand, for z = −2i, |z| = √ (−2) 2 = √ 4 = 2 and also by θ = −π
2
is found.
z = 2(cos( −π ) + i. sin(−π
2 2 ))
3. If z = −5 ⇒ |z| = √ (−5) 2 = √ 25 = 5. As θ = π, we have
z = 5(cos π + i. sin π).
√
4. For z = 1 2 + 1 4
√( πi, we calculate desired result which |z| = 1 2 )2 + ( 1 4 π)2 π
=
2 +4
16
. Besides due to
and
⇒ z =
5. Let z = 1+i
1−i
. Firstly, we have to say that
√
π 2 + 4
16 (cos(arctan(π 2 )) + i. sin(arctan(π 2 ))).
z = 1 + i (1 + i)(1 − i)
=
1 − i (1 − i) 2 = 1 − i + i − i2
1 − 2i + i 2 = 1 − (−1)
1 − 2i − 1 = 2
−2i =
2i
(−2i.i) = 2i
2 = i.
After that, now we could calculate |z| = √ 1 2 = √ 1 = 1. Moreover, θ = π 2
, we conclude that
z = cos( π 2 ) + i. sin(π 2 ).
6. Let z = 3√ 2+2i
− √ 2−2 √ 3i . Then
z =
3√ 2 + 2i
− √ 2 − 2 √ 3i = (3√ 2 + 2i)(− √ 2 + 2 √ 3i)
(− √ 2 − 2 √ 3i)(− √ 2 + 2 √ 3i) = (−6 + 6√ 6i − 2 √ 2i − 4 √ 3)
= −11
−10
5 .
Finally, we have |z| =
√
( −11
5 )2 =
√
121
25 = 11 5
. From θ = π,
7. If z = −6+5i
3i
, we can alternatively interpret it:
z = 11 (cos π + i. sin π).
5
z =
−6 + 5i
3i
=
(−6 + 5i)(−3i)
(3i)(−3i)
=
18i + 15
9
= 5 3 + 2i.
By this reason, we get
|z| =
√
( 5 3 )2 + 2 2 =
√
25
9 + 4 = √
61
3 .
4

As we know that tan θ = 6 5 , we immediately conclude that θ = arctan 6 5 and
⇒ z =
√
61
3 (cos(arctan(6 5 )) + i. sin(arctan(6 5 ))).
8. Let z = 2+3i
5+4i
. This can be simple form as following :
From this equality,
z = 2 + 3i (2 + 3i)(5 − 4i) 22 + 7i
= = == 22
5 + 4i (5 + 4i)(5 − 4i 41 41 + 7 41 i.
|z| =
From above, we can say that
consequently
√
⇒ z =
( 22
41 )2 + ( 7
41 )2 =
√
444
1681 + 49
1681 = √
533
41 .
tan θ = 7 22 ⇒ θ = arctan 7 22 ,
9-15 Determine the principle value of the argument:
√
533
41 (cos(arctan(7 2 )) + i. sin(arctan(7 2 ))).
9. Let z = −1 − i. Then tan α = 1 and since z is in 3. region, we get α = 5π 4 .
10. Let z = −20 + i. For z,
tan α = −1
20 ⇒ α = arctan(−1 ) = 3, 09163
20
On the other hand, for z = −20 − i, tan β = 1
20
and so
⇒ β = arctan( 1 ) = −3, 09163.
20
11. We take z = 4 ∓ 3i. As a result,
tan α = ∓3
4
⇒ α = arctan( ∓3 ) = ∓0, 6435.
4
12. If z = −π 2 , then tan α = 0 and due to this we get α = π.
13. Let z = 7 + 7i. In conclusion
⇒ tan α = 1 ⇒ α = arctan( π ) = 1, 5485.
4
On the other hand, for z = 7 − 7i
tan α = −1 ⇒ α = arctan( −π ) = −1, 5485.
4
14. Take z = (1 + i) 12 . We can this clearly as following :
z = (1 + i) 12 = ((1 + i) 2 ) 6 = (1 + 2i − 1) 6 = 64(i 2 ) 3 = 64(−1) 3 = −64.
5

For this reason, tan α = 0 and so α = π.
15. Let z = (9 + 9i) 3 . This could be interpret more simple:
z = (9 + 9i) 3 = 9 3 + 3.9 2 .9i + 3.9.9 2 .i 2 + 9 3 .i 3 = 729 + 2187i − 2187 − 729i = −1458(1 + i).
By this reality, tan α = 1 and since z is in 3. region, we conclude that α = 5π 4 .
16-20 Represent in the form x + iy:
16. We take z = cos 1 ∓1
2π + i. sin
2 π. Moreover, we know cos 1 2 π = 0 and sin 1 2π = 1. Then we get that
z = cos 1 ∓1
π + i. sin
2 2 π = ∓i.
17. If z = 3(cos 0.2 + i. sin 0.2), then from cos 0.2 = 2, 99 and sin 0.2 = 0, 01,
z = 3(cos 0.2 + i. sin 0.2) = 3. cos 0.2 + 3. sin 0.2 = 2, 99 + 0, 01i
is found.
18. Let z = 4(cos π 3 ∓ i sin π 3
). Because
the result is
cos π 3 = 1 2 and sin π 3 = √
3
2 ,
4(cos π 3 ∓ i sin π 3 ) = 4.(1 2 ∓ i. √
3
2 ) = 2 ∓ 2√ 3i.
19. For z = cos(−1) + i. sin(−1), since cos(−1) = 0, 99 and sin(−1) = −0, 01, we have
cos(−1) + i. sin(−1) = 0, 99 − 0, 01i.
20. Let z = 12(cos 3π 2 + i sin 3π 2
). Because of the fact that
cos 3π 2 = 0 and sin 3π 2 = −1,
we conclude that
12(cos 3π 2 + i sin 3π 2
) = 12.(0 + i.(−1)) = −12i.
21-25 Find all roots in the complex plane:
21. Let z = x + iy. Then
z = √ −i ⇒ z 2 = −i ⇒ x 2 − y 2 + 2xyi = −i ⇒ x 2 − y 2 = 0, 2xy = −1.
As a consequence,
y 4 = 1 4 ⇒ x = y = ∓ 1 √
2
⇒ z = ∓ 1 √
2
∓ 1 √
2
i.
6

22. Let z = 8√ 1. Since
1 = cos(2kπ) + i sin(2kπ), k = 0, ±1, ±2, ...,
Above equation can be rewrite in the following form:
There are 8 roots as following:
23. Let z = x + iy. If z = 4√ −1, then z 4 = −1.
z = 8√ 1 = cos( 2kπ
8 ) + i sin(2kπ 8 )
k = 0 =⇒ z 0 = 1,
k = 1 =⇒ z 1 = 1 √
2
+ i 1 √
2
,
k = 2 =⇒ z 2 = i,
k = 3 =⇒ z 3 = − 1 √
2
+ i 1 √
2
,
k = 4 =⇒ z 4 = −1,
k = 5 =⇒ z 5 = − 1 √
2
− i 1 √
2
,
k = 6 =⇒ z 6 = −i,
k = 7 =⇒ z 7 = 1 √
2
− i 1 √
2
,
(x+iy) 4 = (x 4 −6x 2 y 2 +y 4 )+(4x 3 y−4xy 3 )i = −1 ⇒ x 4 −6x 2 y 2 +y 4 = −1 and 4x 3 y−4xy 3 = 0.
Solving latter equation, we have x = y. After, solving first equation with x = y, we have x = y = ∓ 1 √
2
.
Therefore, z = ∓ 1 √
2
∓ 1 √
2
i.
24. Let w = 3 + 4i and z = 3√ w. From this, we get
arg(w) = α =⇒ tan α = 4 3 =⇒ arctan(4 3 ) = α.
For θ = arg(z), we can say that
=⇒ θ = α 3 = 1 3 arctan(4 3 ).
There are 3 roots:
=⇒ z k = 3√ 5(cos(θ + 2kπ
2kπ
) + i sin(θ +
3 3 ))
k = 0 =⇒ z 0 = 3√ 5(cos θ + i sin θ)
k = 1 =⇒ z 1 = 3√ 5(cos(θ + 2π 3 ) + i sin(θ + 2π 3 ))
7

k = 2 =⇒ z 2 = 3√ 5(cos(θ + 4π 3 ) + i sin(θ + 4π 3 )).
25. Taking z = 5√ −1, z 5 = −1 = cos π + i sin π can be written. So
z k = cos( π + 2kπ
5
) + i sin( π + 2kπ )
5
For k = 0, 1, 2, 3, 4, 5, respectively there are 6 roots. These are as following :
z 0 = cos π 5 + i sin π 5 ,
z 1 = cos 3π 5 + i sin 3π 5 ,
z 2 = −1,
z 3 = cos 7π 5 + i sin 7π 5 ,
z 4 = cos 9π 5 + i sin 9π 5 .
27-30 Solve the following equations:
27. To find all roots of z 2 − (8 − 5i)z + 40 − 20i, we should calculate
∆ = (8 − 5i) 2 − 4.1.(40 − 20i) = −121.
Consequently,
Two roots are z 1 = 4 − 8i and z 2 = 4 + 3i.
⇒ z 1,2 = 8 − 5i ∓ √ −121
2
28. Let’s determine the all roots of z 4 + (5 − 14i)z 2 − (24 + 10i). For this reason, if we take z 2 = u, we
get u 2 + (5 − 14i)u − (24 + 10i) = 0. So we find that
∆ = (5 − 14i) 2 − 4.1.(−24 − 10i) = −85 − 100i.
Hence two roots are following:
u 1,2 = 14 − i ∓ √ −85 − 100i
.
2
29. Given 8z 2 − (36 − 6i)z + 42 − 11i = 0, we now find all roots. First of all,
∆ = (36 − 6i) 2 − 4.8.(42 − 11i) = −84 + 96i
is calculated. By this knowledge, we say
⇒ z 1,2 = 36 − 6i ∓ √ −84 + 96i
.
16
8

30. For z 4 + 16 = 0, let z 2 = u. Then
√
i can be determined as following:
u 2 = −16 ⇒ u = ∓4i ⇒ z 2 = ∓4i ⇒ z = ∓2 √ i.
As a result, all roots are ∓(1 ∓ i). √ 2.
√
i = ∓[
√
1
2 + 1.i. √
1
2 ] = ∓ 1 √
2
(1 + i).
13.3
1-10 Find and graph followings in the complex plane:
1. Let |z − 3 − 2i| = 4 3
and z = a + ib. Since
|z − 3 − 2i| = |(a − 3) + (b − 2)i| = √ (a − 3) 2 + (b − 2) 2 = 4 3 ⇒ (a − 3)2 + (b − 2) 2 = 16
9 ,
we get a circle with radius r = 4 3
and centered at (3, 2). For graph, look at the following picture.
2. Let 1 ≤ |z − 1 + 4i| ≤ 5 and z = a + bi. Because of
1 ≤ |z − 1 + 4i| ≤ 5 ⇒ 1 ≤ |(a − 1) + (b + 4)i| ≤ 5 ⇒ 1 ≤ (a − 1) 2 + (b + 4) 2 ≤ 25,
we find that closed annulus bounded by circles of radius 1 and 5 centered at (1, −4).
3. If we take 0 < |z − 1| < 1 and z = a + bi, then we get
0 < |z − 1| < 1 ⇒ 0 < |(a − 1) + bi| < 1 ⇒ 0 < (a − 1) 2 + b 2 < 1,
i.e. a circle with radius 1 and centered at (1, 0).
9

4. Let’s now determine the region of −π < Rez < π. For this reason, we take z = x + iy. As
−π < Rez < π ⇒ −π < x < π,
we get open vertical strip of width 2π. For the region, look at this picture.
5. Let Imz 2 = 2 and z = x + iy. From z 2 = (x + iy) 2 = x 2 − y 2 + 2xyi, we can say that
Im(x 2 − y 2 + 2xyi) = 2 ⇒ 2xy = 2 ⇒ xy = 1.
Consequently, its region is pictured as following:
6. For Rez > −1, let z = x + iy. Then
Rez = x > 1, y ∈ R.
So, open half-plane extending from the vertical line x = 1 to the right is found.
10

7. Let |z + 1| = |z − 1| and z = a + bi. If same things occur,
|z + 1| = |z − 1| ⇒ |(a + 1) + bi| = |(a − 1) + bi| ⇒ √ (a + 1) 2 + b 2 = √ (a − 1) 2 + b 2
⇒ (a + 1) 2 + b 2 = (a − 1) 2 + b 2
⇒ a = 0 ⇒ z = bi, b ∈ R
is found. Then we could picture it as following:
8. We now determine the region of |Argz| ≤ 1 4π. For this reason, z = x + iy is taken.
|Argz| ≤ 1 4 π ⇒ | tan( y x )| ≤ 1 4 π ⇒ −π
4 ≤ tan( y x ) ≤ π 4
⇒ arctan( −π
4 ) ≤ y x ≤ arctan(π 4 ) ⇒ −1 ≤ y x ≤ 1.
Angular region of angle π 2
symmetric to the positive x-axis is found.
9. If Rez ≤ Imz and z = x + iy, then x ≤ y. So
10. For Re( 1 z
) < 1, we take z = x + iy. Because we know that
we conclude that
1
Re(
x + iy ) ⇒
1
z = 1
x + iy =
This is the exterior of the circle of radius 1 2 centered at 1 2 .
x
x 2 + y 2 ,
x
x 2 + y 2 < 1 ⇒ 1 4 < (x − 1 2 )2 + y 2 .
11

12-15 Function values:
12. For z = 2 + i, since f(z) = 3z 2 − 6z + 3i,
f(z) = f(2 + i) = 3(2 + i) 2 − 6(2 + i) + 3i = 3(4 + 4i − 1) − 12 − 6i + 3i
= 3(3 + 4i) − 12 − 3i = 9 + 12i − 12 − 3i = −3 + 9i
is found. Let z = x + iy.
f(z) = 3(x + iy) 2 − 6(x + iy) + 3i = (3x 2 − 3y 2 − 6x) + (6xy − 6y + 3)i
13. Let z = 4 − 5i. Because f(z) = z
z+1
, we get
⇒ Ref = 3x 2 − 3y 2 − 6x, Imf = 6xy − 6y + 3.
f(z) = f(4 − 5i) = 4 − 5i (4 − 5i)(5 + 5i) 20 + 20i − 25i + 25 45 − 5i
= = = = 9 − i
5 − 5i (5 − 5i)(5 + 5i) 25 + 25
50 10 .
Let z = x + iy. Then
f(z) =
x + iy
x + 1 + iy =
(x + iy)(x + 1 − iy)
(x + 1 + iy)(x + 1 − iy) = x2 + x + y 2 + (−xy + y)i
x 2 + 2x + 1 + y 2
⇒ Ref = x2 + x + y 2
x 2 + 2x + 1 + y 2 , Imf =
14. Let z = 1 2 + 1 1
4i. Since f(z) =
1−z
, we have
Let z = x + iy.
−xy + y
x 2 + 2x + 1 + y 2 .
f(z) = f( 1 2 + 1 4 i) = 1
1 − 1 2 − 1 4 i = 1
1
1
2 − 1 4 i = 2 + 1 4 i
( 1 2 − 1 4 i)( 1 2 + 1 = 1, 6 + 0, 8i.
4i) f(z) =
1
1 − x − iy = 1 − x + iy
(1 − x − iy)(1 − x + iy) = ( 1 − x
(1 − x) 2 + y 2 ) + ( y
(1 − x) 2 + y 2 )i
As a result, we have
⇒ Ref =
1 − x
(1 − x) 2 + y 2 , Imf = y
(1 − x) 2 + y 2 .
12

15. Let z = 1 + i. Since f(z) = 1
z 2 , we have
For z = x + iy,
f(z) =
f(z) = f(1 + i) =
1
(x + iy) 2 = 1
x 2 − y 2 + 2xyi =
and so we can say
16-19 Continuity:
Ref =
1
(1 + i) 2 = 1
1 + 2i − 1 = 1 2i = −2i
(2i)(−2i) = −2i
4 = − i 2 .
x 2 − y 2 − 2xyi
(x 2 − y 2 + 2xyi)(x 2 − y 2 − 2xyi) = x2 − y 2 − 2xyi
(x 2 − y 2 ) 2 + 4x 2 y 2
x 2 − y 2
(x 2 − y 2 ) 2 + 4x 2 y 2 , Imf = −2xyi
(x 2 − y 2 ) 2 + 4x 2 y 2 .
16. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
[Re(z 2 )]/|z| 2 = r 2 . cos 2θ/r 2 = cos 2θ,
this function isn’t continuous at z = 0.
17. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
[Im(z 2 )]/|z| = r 2 . sin 2θ/r = r. sin 2θ r→0 −→ 0,
this function is continuous at z = 0.
18. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ) and Re( 1 z
|z| 2 .Re( 1 z ) = r2 .r. cos θ/r 2 = r. cos θ r→0 −→ 0,
r. cos θ
) = . Because
r 2
this function is continuous at z = 0.
19. Let z = r.(cos θ + i. sin θ). Then z 2 = r 2 .(cos 2θ + i. sin 2θ). Because
this function is continuous at z = 0.
20-24 Derivative:
(Imz)/(1 − |z|) = r. sin θ/(1 − r) =
20. ( z2 −9
z 2 +1 )′ = (2z).(z2 +1)−(z 2 −9).(2z)
(z 2 +1) 2 = 2z3 +2z−2z 3 +18z
(z 2 +1) 2 = 20z
(z 2 +1) 2 .
21. [(z 3 + i) 2 ] ′ = 2.(z 3 + i).3z 2 = 6z 5 + 6z 2 i.
22. ( 3z+4i
1,5iz−2 )′ = 3.(1,5iz−2)−(3z+4i).1,5i = 4,5iz−6−4,5iz+6 = 0.
(1,5iz−2) 2
(1,5iz−2) 2
23. (
i
(1−z) 2 ) ′ = −i.2.(1−z).(−1)
(1−z) 4 = 2i
(1−z) 3 .
24. (
z 2
(z+i) 2 ) ′ = 2z.(z+i)2 −z 2 .2.(z+i)
(z+i) 4 = 2zi
(z+i) 3 .
r r→0
. sin θ −→ 0,
1 − r
13

13.4
1-10 Cauchy-Riemann equations:
1. For f(z) = z 4 , let z = x + iy.
=⇒ (x + iy) 4 = x 4 + 4x 3 iy + 6x 2 i 2 y 2 + 4xi 3 y 3 + i 4 y 4
= (x 4 − 6x 2 y 2 + y 4 ) + (4x 3 y − 4xy 3 )i
Let u = x 4 − 6x 2 y 2 + y 4 and v = 4x 3 y − 4xy 3 .
u x = 4x 3 − 12xy 2 , u y = −12x 2 y + 4y 3 , v x = 12x 2 y − 4y 3 and v y = 4x 3 − 12xy 2 .
=⇒ u x = v y and u y = −v x =⇒ f is analytic.
2. f(z) = Im(z 2 )
z 2 = x 2 − y 2 + 2xyi =⇒ f(z) = 2xy.
Let u = 2xy and v = 0. =⇒ u x = 2y, u y = 2x, v x = 0 and v y = 0.
=⇒ u x ≠ v y and u y ≠ −v x =⇒ f isn’t analytic.
3. e 2x .(cos y + i sin y) =⇒ Let be u = e 2x cos y and v = e 2x sin y
u x = 2e 2x cos y, u y = −e 2x sin y, v x = 2e 2x sin y and v y = e 2x cos y.
=⇒ u x ≠ v y , u y ≠ −v x =⇒ f isn’t analytic.
4. f(z) = 1
1−z 4 , 1 − z 4 ≠ 0 =⇒ z 4 ≠ 1 =⇒ When z ≠ ∓i and z ≠ ∓1, f is analytic.
5. e −x (cos y − i sin y)
Let u = e −x cos y and v = e −x sin y.
=⇒ u x = −e −x cos y, u y = −e −x sin y, v x = e −x sin y and v y = −e −x cos y.
=⇒ u x = v y and u y = −v x =⇒ f is analytic.
6. f(z) = arg(πz)
πz = πx + iπy =⇒ arg(πz) = arctan( y x )
=⇒ u = arctan( y x ) and v = 0 =⇒ u x = x2
x 2 +y 2
7. f(z) = Rez + Imz = x + y =⇒ u = x + y and v = 0.
=⇒ u x = 1 ≠ v y = 0 =⇒ f isn’t analytic function.
8. f(z) = ln |z| + i. arg z, |z| = √ x 2 + y 2
≠ v y = 0 =⇒ f isn’t analytic.
f is analytic for every z ∈ C, |z| > 0 =⇒ ln |z| is defined =⇒ arg z = θ is defined. =⇒ f is analytic.
9. f(z) = i
z 8
When z ≠ 0, f is analytic.
10. f(z) = z 2 + 1 = z4 +1
z 2 z 2
When z ≠ 0, f is analytic.
11. Let x = r cos θ, y = r sin θ and u x = v y , u y = −v x .
∂u
∂r = ∂u
∂x .∂x ∂r + ∂u
∂y .∂y ∂r = u x cos θ + u y sin θ = v y cos θ − v x sin θ
14

∂u
∂θ = ∂u
∂x .∂x ∂θ + ∂u
∂y .∂y ∂θ = u x(−r sin θ) + u y (r cos θ) = −rv y sin θ − rv x cos θ
12-21 Harmonic functions:
12. u = xy
∂v
∂r = ∂v
∂x .∂x ∂r + ∂v
∂y .∂y ∂r = v x cos θ + v y sin θ
∂v
∂θ = ∂v
∂x .∂x ∂θ + ∂v
∂y .∂y ∂θ = v x(−r sin θ) + v y (r cos θ)
=⇒ u r = 1 r v θ and v r = − 1 r u θ.
u xx = 0 and u yy = 0 =⇒ u xx + u yy = 0 =⇒ u is harmonic.
u x = v y =⇒ y = v y =⇒ v = y2
2 + h(x) =⇒ v x = dh
dx
u y = −v x =⇒ x = − dh
x2
y2
dx
=⇒ h(x) = −
2
+ c =⇒ f(z) = xy + (
2 − x2
2 + c)i.
13. v = xy
v xx = 0, v yy = 0 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
v x = −u y =⇒ u y = −y =⇒ u = − y2
2 + h(x) =⇒ u x = dh
dx .
Since u x = v y = x, dh
x2
dx
= x =⇒ h(x) =
2 + c.
=⇒ f(z) = − y2
2 + x2
2 + c + xyi.
14. v = − y
x 2 +y 2
−u y = v x =
2xy , u
(x 2 +y 2 ) 2 x = v y = y2 −x 2
(x 2 +y 2 ) 2
=⇒ v xx = −6x2 y+2y 3
(x 2 +y 2 ) 3 and v yy = 6x2 y−2y 3
(x 2 +y 2 ) 3 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
u =
x
x 2 +y 2 + c =⇒ f(z) =
15. u = ln |z| = ln √ x 2 + y 2
v y = u x =
x
2(x 2 +y 2 ) , −v x = u y =
x
x 2 +y 2 + c + i
y
2(x 2 +y 2 )
−y
.
x 2 +y 2
=⇒ u xx = y2 −x 2
2(x 2 +y 2 ) 2 and u yy = x2 −y 2
2(x 2 +y 2 ) 2 =⇒ u xx + u yy = 0 =⇒ u is harmonic.
v = 1 2 arctan( y x ) + h(x) =⇒ v x = −
=⇒ f(z) = ln |z| + 1 2 arctan( y x ) + c.
16. v = ln |z| = ln √ x 2 + y 2
−u y = v x =
x
2(x 2 +y 2 ) , u x = v y =
y
2(x 2 +y 2 ) + dh
dx
y
2(x 2 +y 2 )
dh
=⇒
dx
= 0 =⇒ h(x) = c.
=⇒ v xx = y2 −x 2
2(x 2 +y 2 ) 2 and v yy = x2 −y 2
2(x 2 +y 2 ) 2 =⇒ v xx + v yy = 0 =⇒ v is harmonic.
u = −1
2 arctan( y x ) + h(x) =⇒ u x =
=⇒ f(z) = −1
2 arctan( y x
) + c + i ln |z|.
17. u = x 3 − 3xy 2
v y = u x = 3x 2 − 3y 2 , −v x = u y = −6xy
y
2(x 2 +y 2 ) + dh
dx
=⇒ h(x) = c.
=⇒ u xx = 6x and u yy = −6x =⇒ u xx + u yy = 0 =⇒ u is harmonic.
v = 3x 2 y − y 3 + h(x) =⇒ v x = 6xy + dh
dx
=⇒ h(x) = c.
15

=⇒ f(z) = x 3 − 3xy 2 + i(3x 2 y − y 3 + c).
18. u = 1
x 2 +y 2
u x = −
2x , u
(x 2 +y 2 ) 2 y = −2y
(x 2 +y 2 ) 2
=⇒ u xx = 6x2 −2y 2
(x 2 +y 2 ) 3 and u yy = 6y2 −2x 2
(x 2 +y 2 ) 3 =⇒ u xx + u yy ≠ 0 =⇒ u isn’t harmonic.
19. v = (x 2 − y 2 ) 2
−u y = v x = 4x 3 − 4xy 2 , u x = v y = −4x 2 y + 4y 3
=⇒ v xx = 12x 2 −4y 2 and v yy = −4x 2 +12y 2 =⇒ When x, y ≠ 0v xx +v yy ≠ 0 =⇒ v isn’t harmonic.
20. u = cos x cosh y
v y = u x = − sin x cosh y, −v x = u y = cos x sinh y
=⇒ u xx = − cos x cosh y and u yy = cos x cosh y =⇒ u xx + u yy = 0 =⇒ u is harmonic.
v = − sin x sinh y + h(x) =⇒ v x = − cos x sinh y + dh
dx
=⇒ h(x) = c.
=⇒ f(z) = cos x cosh y − i sin x sinh y + c.
21. u = e −x sin 2y
u x = −e −x sin 2y, u y = 2e −x cos 2y
=⇒ u xx = e −x sin 2y and u yy = −4e −x sin 2y =⇒ When y ≠ kπ, u xx + u yy = −3e −x sin 2y ≠ 0
=⇒ u isn’t harmonic.
22-24 Harmonic conjugate:
22. u = e 3x cos ay harmonic =⇒ u xx + u yy = 0.
u x = 3e 3x cos ay, u y = −ae 3x sin ay
=⇒ u xx = 9e 3x cos ay and u yy = −a 2 e 3x cos ay =⇒ u xx +u yy = e 3x cos ay(9−a 2 ) = 0 =⇒ a = ∓3
When a = −3, u = e 3x cos(−3y)
v y = u x = 3e 3x cos(−3y), −v x = u y = 3e 3x sin(−3y) =⇒ v = −e 3x sin(−3y) + h(x)
=⇒ v x = −3e 3x sin(−3y) + dh
dx
=⇒ h(x) = c.
=⇒ v = −e 3x sin(−3y) + c.
When a = 3, u = e 3x cos(3y)
v y = u x = 3e 3x cos(3y), −v x = u y = −3e 3x sin(3y) =⇒ v = e 3x sin(3y) + h(x)
=⇒ v x = 3e 3x sin(3y) + dh
dx
=⇒ h(x) = c.
v = e 3x sin(3y) + c
23. u = sin x cosh(cy) harmonic =⇒ u xx + u yy = 0.
u x = cos x cosh(cy), u y = c sin x sinh(cy)
=⇒ u xx = −sinx cosh(cy) and u yy = c 2 sin x cosh(cy) and u xx + u yy = 0 =⇒ c 2 − 1 = 0
=⇒ c = ∓1
When c = −1, u = sin x cosh(−y)
16

v y = u x = cos x cosh(−y), −v x = u y = c − sin x sinh(−y) =⇒ v = − cos x sinh(−y) + h(x)
=⇒ v x = sin x sinh(−y) + dh
dx
=⇒ h(x) = c
=⇒ v = − cos x sinh(−y) + c.
When c = 1, u = sin x cosh(y)
v y = u x = cos x cosh(y), −v x = u y = c sin x sinh(y) =⇒ v = cos x sinh(y) + h(x)
=⇒ v x = − sin x sinh(y) + dh
dx
=⇒ h(x) = c
=⇒ v = cos x sinh(y) + c.
24. u = ax 3 + by 3 harmonic =⇒ u xx + u yy = 0.
u x = 3ax 2 , u y = 3by 2
=⇒ u xx = 6ax and u yy = 6by =⇒ u xx + u yy = 0 and ax + by = 0 =⇒ a = b = 0.
=⇒ u x = v y = 0 =⇒ v = h(x) =⇒ v x = dh
dx
=⇒ h(x) = c =⇒ v = c is constant.
13.5
1. e z = e x+iy = e x .e iy = e x (cos y + i sin y)
u = e x cos y and v = e x sin y =⇒ u x = e x cos y, u y = −e x sin y, v x = e x sin y and v y = e x cos y.
=⇒ u x = v y and u y = −v x .
=⇒ e z is analytic for every z. =⇒ e z is entire.
2-8 Values of e z :
2. z = 3 + πi ⇒ e 3+πi = e 3 (cos π − i. sin π) = −e 3 ∼ = −20, 086 and |e 3+πi | ∼ = 20, 086.
3. e 1+2i = e(cos 2 + i. sin 2) and |e 1+2i | = e √ cos 2 2 + i. sin 2 2.
4. e √ 2− π 2 i = e √2 [cos( −π
−π
2
) + i. sin(
2 )] = −e√2 i ∼ = −4, 11325i and |e √ 2− π 2 i | ∼ = 4, 11325.
5. e 7πi/2 = e 0 [cos(7πi/2) + i. sin(7πi/2)] = i and |e 7πi/2 | = |i| = 1.
6. e (1+i)π = e π (cos π + i sin π) = −e π ∼ = −23, 1407 and |e (1+i)π | ∼ = 23, 1407.
7. e 0,8−5i = e 0,8 .(cos(−5) + i. sin(−5)) = 2, 23.(0, 28 + i.0, 95) and |e 0,8−5i | ∼ = 2, 2.
8. e 9πi/2 = e 0 [cos(9πi/2) + i. sin(9πi/2)] = i and |e 9πi/2 | = |i| = 1.
9-12 Real and Imaginary parts: Let z = x + iy.
9. e −2z = e −2x−2yi = e −2x (cos(−2y) + i. sin(−2y))
⇒ Re(e −2z ) = e −2x (cos(2y) and Im(e −2z ) = −e −2x (sin(2y).
10. e z3 = e (x+iy)3 = e (x3 −3xy 2 )+(3x 2 y−y 3 )i = e (x3 −3xy 2) (cos(3x 2 y − y 3 ) + i. sin(3x 2 y − y 3 ))
⇒ Re(e z3 ) = e (x3 −3xy 2) cos(3x 2 y − y 3 ) and Im(e z3 ) = e (x3 −3xy 2) sin(3x 2 y − y 3 ).
11. e z2 = e (x+iy)2 = e x2 −y 2 +2xyi = e x2 −y 2 (cos(2xy) + i. sin(2xy)).
12. e 1 z = e 1
x+iy
= e x−iy
x 2 +y 2 = e x
x 2 +y 2 [cos(
y
) − i. sin( )].
x 2 +y 2 x 2 +y 2
17
y

13-17 Polar form:
13. z = √ i ⇒ √ √ √
1
i = ∓[
2 + i. 1
2 ] = ∓ √ 1
2
(1 + i) ⇒ |z| = 1, tan θ = 1 ⇒ θ = π 4 , 7π 4
⇒ e π 4 and e 7π 4 .
14. z = 1 + i ⇒ r = |z| = √ 2, tan θ = 1 and because z is in the 1. region, θ = π 4 , so that √ 2e πi
4 .
15. z = r.e iθ ⇒ n√ z = n√ r.e iθ = r 1 n .e iθ n .
16. z = 3 + 4i ⇒ r = |z| = 5 and tan θ = 4 3 ⇒ θ = arctan( 4 3 ), so that 5ei. arctan( 4 3 ) .
17. z = −9 ⇒ r = |z| = 9, because z is on the left of x-axis, θ = π, thus 9e iπ .
18-21 Solution of equations: Let z = x + iy.
18. e 3z = 4 ⇒ e 3x (cos 3y + i. sin 3y) = 4 ⇒ e 3x cos 3y = 4 and e 3x sin 3y = 0
e 6x cos 2 3y + e 6x sin 2 3y = 16 ⇒ e 6x (cos 2 3y + sin 2 3y) = 16 ⇒ e 3x = 4 ⇒ x = ln 4
3
Thus, we get that z = ln 4
3
+ i.
2kπ
3 .
tan 3y = 0 ⇒ y = 2kπ
3 , k ∈ Z
19. e z = −2 ⇒ e x (cos y + i. sin y) = −2 ⇒ e x cos y = −2 and e x sin y = 0
e 2x cos 2 y + e 2x sin 2 y = 4 ⇒ e 2x (cos 2 y + sin 2 y) = 4 ⇒ e 2x = 4 ⇒ x = ln 2
So, we have z = ln 2 + i.(2k − 1)π.
tan y = 0 ⇒ y = (2k − 1)π, k ∈ Z
20. e z = 0 ⇒ e x+iy = 0 ⇒ e x (cos y + i. sin y) = 0 ⇒ e x cos y = e x sin y = 0.
We know that cos y = sin y = 0 isn’t possible at the same time. Hence there is no solution of e z = 0.
21. e z = 4 − 3i ⇒ e x (cos y + i. sin y) = 4 − 3i ⇒ e x cos y = 4 and e x sin y = −3
e 2x cos 2 y + e 2x sin 2 y = 16 ⇒ e 2x (cos 2 y + sin 2 y) = 9 ⇒ e 2x = 25 ⇒ x = ln 5
tan y = −3
4
Finally, we have z = ln 5 − i. arctan( 4 3 ).
⇒ y = − arctan( 3 4 )
1.
13.6
cos z = eiz + e −iz
2
⇒ d dz cos z = i 2 (eiz − e −iz ) = − sin z
cos z is defined and differentiable at all points of C. So, cos z is entire.
sin z = eiz − e −iz
2i
⇒ d dz sinz = ieiz + ie −iz
2i
= eiz + e −iz
2
= cos z
18

sin z is defined and differentiable at all points of C. Hence sin z is entire.
cosh z = ez + e −z
⇒ d 2 dz coshz = ez − e −z
= sinh z
2
cosh z is defined and differentiable at all points of C. So cosh z is entire.
sinh z = ez − e −z
⇒ d 2 dz sinhz = ez + e −z
= cosh z
2
sinh z is defined and differentiable at all points of C. Therefore sinh z is entire.
2. Let z = x + iy. Then we can write
cos z = cos x cosh y − i sin x sinh y
Re(cos z) = cos x cosh y. Let u(x, y) = cos x cosh y and v(x, y) = 0.
u x = − sin x cosh y,
u y = cos x sinh y,
u xx = − cos x cosh y
u yy = cos x cosh y
Then we obtain that ∇ 2 u = u xx + u yy = − cos x cosh y + cos x cosh y = 0.
v xx , v yy = 0 because v(x, y) = 0. So, ∇ 2 v = v xx + v yy = 0.
Hence Re(cos z) is harmonic.
We know that
sin z = sin x cosh y + i cos x sinh y
Then Im(sin z) = cos x sinh y. Let u(x, y) = cos x sinh y and v(x, y) = 0.
u x = − sin x sinh y,
u y = cos x cosh y,
u xx = − cos x sinh y
u yy = cos x sinh y
∇ 2 u = u xx + u yy = − cos x sinh y + cos x sinh y = 0.
Because of v(x, y) = 0, v xx and v yy is zero. Hence ∇ 2 v = v xx + v yy = 0. Im(sin z) is harmonic.
3. We know that cosiz = coshz and siniz = isinhz. Furthermore cos z = cos x cosh y − i sin x sinh y.
cosh z = cos iz = cos(i(x + iy)) = cos(−y + ix) = cos(−y) cosh(x) − i sin(−y) sinh(x)
= cos y cosh x + i sin y sinh x
We know that sin z = sin x cosh y + i cos x sinh y Then,
sinh z =
sin iz
i
= −i sin(iz) = −i sin(−y + ix) = −i(sin(−y) cosh(x) + i cos(−y) sinh(x))
= −i(− sin y cosh x + i cos y sinh x) = cos y sinh x + i sin y cosh x.
4. We know that cos iz = cosh z and sin iz = i sinh z. If we use
cos(z 1 + z 2 ) = cos z 1 cos z 2 − sin z 1 sin z 2
19

then,
cosh(z 1 + z 2 ) = cos(i(z 1 + z 2 ))
= cos(iz 1 + iz 2 )
= cos(iz 1 ) cos(iz 2 ) − sin(iz 1 ) sin(iz 2 )
= cosh z 1 cosh z 2 − i sinh z 1 .i sinh z 2
= cosh z 1 cosh z 2 + sinh z 1 sinh z 2 .
If we use
sin(z 1 + z 2 ) = sin z 1 cos z 2 + sin z 2 cos z 2
then,
sinh(z 1 + z 2 ) = sin(i(z 1+z 2 ))
i
= −i(sin(iz 1 + iz 2 ))
= −i[sin(iz 1 ) cos(iz 2 ) + sin(iz 2 ) cos(iz 1 )]
= −i[i sinh z 1 cosh z 2 + i sinh z 2 . cosh z 1 ]
= sinh z 1 cosh z 2 + cosh z 1 sinh z 2 .
5. cosh 2 z − sinh 2 z = ( ez +e −z
2
) 2 − ( ez −e −z
2
) 2 = e2z +e −2z +2−e 2z −e −2z +2
4
= 4 4 = 1
6. cosh 2 z + sinh 2 z = ( ez +e −z
2
) 2 + ( ez −e −z
2
) 2 = e2z +e −2z +2+e 2z +e −2z −2
4
= e2z+e−2z
2
= cosh 2z
7-15: Function Values
7. cos(1 + i) = ei(1+i) +e −i(1+i)
2
= e−1+i +e 1−i
2
= e−1 (cos 1+i sin 1)+e(cos(−1)+i sin(−1))
2
= e−1 cos 1+e −1 i sin 1+e cos 1−ei sin 1
2
= ( e−1 +e 1
2
). cos 1 + i( e−1 −e
2
). sin 1
= cos1.cos1 + i.(− sin 1). sin 1
= cos 2 1 − i sin 2 1.
8. sin(1 + i) = ei(1+i) −e −i(1+i)
2i
= e−1+i −e 1−i
2i
= e−1 (cos 1+i sin 1)−e 1 (cos(−1)+i sin(−1))
2i
= cos 1( e−1 −e
2i
) + i sin 1( e−1 +e
2i
)
= cos 1.(− sin 1) + i sin 1. cos 1
i
= − cos 1 sin 1 + cos 1 sin 1
= 0
9. sin 5i = ei(5i) −e −i(5i)
2i
= e−5 −e 5
2i
= −2i.(e−5 −e 5 )
4
= −i.(e−5 −e 5 )
2
= −i(− sinh 5) = i sinh 5.
20

10. cos 3πi = ei(3πi) +e −i(3πi)
2
= e−3π +e 3π
2
= cosh 3π.
11. cosh(−2 + 3i) = e−2+3i +e 2−3i
2
= e−2 (cos 3+i sin 3)+e 2 (cos(−3)+isin(−3))
2
= cos 3( e−2 +e 2
2
) + i sin 3( e−2 −e 2
2
)
= cos 3 cosh 2 − i sin 3sinh2.
12. • −i sinh(−π + 2i) = −i( e−π+2i −e π−2i
2
)
• sin(2 + πi) =
= −i( e−π (cos 2+i sin 2)−e π (cos(−2)+i sin(−2))
2
)
= −i(cos 2( e−π −e π
2
) + i sin 2( e−π +e π
2
))
= −i(cos 2(− sinh π) + i sin 2 cosh π)
= sin 2 cosh π + i cos 2 sinh π.
sinh i(2+πi)
i
= sinh(−π+2i)
i
13. cosh(2n + 1)πi = e(2n+1)πi +e −(2n+1)πi
2
=
−i sinh(−π+2i)
1
= i cos 2 sinh π + sin 2 cosh π.
= e0 (cos((2n+1)π)+i sin((2n+1)π))+e 0 (cos((−2n−1)π)+i sin((−2n−1)π)
2
= cos((2n+1)π)+cos((−2n−1)π)
2
= cos(π)+cos(−π)
2
= −1, n = 1, 2, ...
14. sinh(4 − 3i) = e4−3i −e −4+3i
2
= e4 (cos(−3)+i sin(−3))−e −4 (cos 3+i sin 3)
2
= cos 3( e4 −e −4
2
) + i sin 3( −e4 −e −4
2
)
= cos 3 sinh 4 − i sin 3 cosh 4.
15. cosh(4 − 6πi) = e4−6πi −e −4+6πi
2
= e4 (cos(−6π)+i sin(−6π))+e −4 (cos(6π)+i sin(6π))
2
= e4 +e −4
2
= cosh 4.
16. We know that tan a = sin a
cos a
tan a+tan b
and tan(a + b) =
1−tan a tan b .
tan z = tan(x + iy) =
tan x+tan iy
1−tan x tan iy =
sin x sin iy
+ cos x cos iy
1− sin x
cos x
.
sin iy
cos iy
=
sin x cos iy+sin iy cos x
cos x cos iy
cos x cos iy−sin x sin iy
cos x cos iy
=
sin x cos iy+sin iy cos x
cos x cos iy−sin x sin iy
=
sin x cosh y+i sinh y cos x
cos x cosh y−i sin x sinh y = sin x cos x cosh2 y+i cos 2 x cosh y sinh y+i sin 2 x sinh y cosh y−sinh 2 y cos x sin x
cos 2 x cosh 2 y+sin 2 x sinh 2 y
= cosh y sinh y(i(cos2 x+sin 2 x))+cos x sin x(cosh 2 y−sinh 2 y)
cos 2 x(1+sinh 2 y)+sin 2 x sinh 2 y
=
i cosh y sinh y+cos x sin x
cos 2 x+cos 2 x sinh 2 y+sin 2 x sinh 2 y
=
cos x sin x+i cosh y sinh y
cos 2 x+sinh 2 y(cos 2 x+sin 2 x) =
Hence we get Re(tan z) =
17-21: Equations
cos x sin x cosh y sinh y
+ i
cos 2 x+sinh 2 y
cos x sin x
and Im(tan z) =
cos 2 x+sinh 2 y
17. 0 = cosh z = ez +e −z
2
= 1 2 (ez + 1
e
) = 1 e 2z +1
z 2 e
. z
cos 2 x+sinh 2 y
cosh y sinh y
cos 2 x+sinh 2 y .
21

Then we obtain that
e 2z + 1 = 0 ⇒ e 2z = −1 ⇒ e 2z = (i) 2 ⇒ e z = i ⇒ ln e z = ln i ⇒ z = ln i.
We know that ln z = ln |z| + i(Arg(z) + 2kπ), k ∈ Z. Then,
z k = ln i = ln |i| + i(Arg(i) + 2kπ) = ln 1 + i( π 2 + 2kπ) = i(π 2 + 2kπ)
So, z k = i (4k+1)π
2
, k ∈ Z.
18. 100 = sin z = eiz −e −iz
2i
Let t = e iz . Then we can write t 2 − 200it − 1 = 0.
⇒ 200i = e2iz −1
e iz ⇒ 200i.e iz = e 2iz − 1 ⇒ e 2iz − 200i.e iz − 1 = 0.
△ = b 2 − 4ac, a = 1, b = −200i, c = −1 ⇒ △ = −39996
t 1,2 = −b+√ △
2a
= 200i+199,9i
2
⇒ t 1 = 399,9i
2
and t 2 = 0,1i
2
• t 1 = e iz 1
⇒ ln t 1 = ln e iz 1
= iz 1 ⇒ z 1 = 1 i [ln |t 1| + i(Arg(t 1 ) + 2kπ)]
= −i[ln( 399,9
2
) + i( π 399,9
2
+ 2kπ)] = −i ln(
2
) + 4k+1
2
π, k ∈ Z
• t 2 = e iz 2
⇒ ln t 2 = ln e iz 2
= iz 2 ⇒ z 2 = 1 i [ln |t 2| + i(Arg(t 2 ) + 2kπ)]
= −i[ln( 1
20 ) + i( π 1
2
+ 2kπ)] = −i ln(
20 ) + 4k+1
2
π, k ∈ Z
19. 2i = cos z = eiz +e −iz
2
= 1 2 ( e2iz +1
e iz ) ⇒ e 2iz − 4ie iz + 1 = 0
Let t = e iz . Then we obtain t 2 − 4it + 1 = 0.
△ = b 2 − 4ac, a = 1, b = −4i, c = 1 ⇒ △ = −20
t 1,2 = −b+√ △
2a
= 4i+2√ 5i
2
= i(2+ √ 5) ⇒ t 1 = i(2 + √ 5) and t 2 = i(2 − √ 5)
• t 1 = e iz 1
⇒ ln t 1 = ln e iz 1
= iz 1 ⇒ z 1 = 1 i [ln |t 1| + i(Arg(t 1 ) + 2kπ)]
= −i[ln(2 + √ 5) + i( π 2 + 2kπ)] = −i ln(2 + √ 5) + 4k+1
2
π, k ∈ Z
• t 2 = e iz 2
⇒ ln t 2 = ln e iz 2
= iz 2 ⇒ z 2 = 1 i [ln |t 2| + i(Arg(t 2 ) + 2kπ)]
= −i[ln(2 − √ 5) + i(− π 2 + 2kπ)] = −i ln(√ 5 − 2) + 4k−1
2
π, k ∈ Z
20. −1 = cosh z = ez +e −z
2
⇒ e 2z + 2e z + 1 = 0
Let t = e z . Then we obtain
t 2 + 2t + 1 = 0 ⇒ (t + 1) 2 = 0 ⇒ (t + 1) = 0 ⇒ t = −1
22

e z = t = −1 ⇒ ln e z = ln(−1)
⇒ z = ln(−1) = ln | − 1| + i(Arg(−1) + 2kπ) k ∈ Z
⇒ z = ln 1 + i(2k + 1)π, k ∈ Z
21. 0 = sinh z = ez −e −z
2
⇒ e z − e −z = 0 ⇒ e 2z = 1
ln e 2z = ln 1 ⇒ 2z = ln |1| + i(Arg(1) + 2kπ), k ∈ Z
⇒ 2z = ln 1 + i(0 + 2kπ), k ∈ Z
⇒ z = kπi, k ∈ Z.
13.7
1-9 Find Ln(z) when z equals:
1. Ln(−10) = ln | − 10| + i.Arg(−10) = ln 10 + πi.
2. Ln(2 + 2i) = ln |2 + 2i| + i.Arg(2 + 2i) = ln( √ 2 2 + 2 2 ) + i. arctan( 2 2 ) = ln(√ 8) + i. arctan(1) =
1
2 ln 8 + i. π 4 .
3. Ln(2 − 2i) = ln |2 − 2i| + i.Arg(2 − 2i) = 1 2 ln 8 + i. 7π 4 .
4. Ln(−5 ∓ 0.1i) = ln | − 5 ∓ 0.1i| + i.Arg(−5 ∓ 0.1i) = ln(5.001) ∓ i.0.02
5. Ln(−3 − 4i) = ln | − 3 − 4i| + i.Arg(−3 − 4i) = ln 5 + i. arctan( 4 3 ).
6. Ln(−100) = ln | − 100| + i.Arg(−100) = ln 100 + i.π = 4.605 + 3.142i.
7. Ln(0.6 + 0.8i) = ln |0.6 + 0.8i| + i.Arg(0.6 + 0.8i) = ln 1 + i. arctan( 4 3 ) = arctan( 4 3 )i.
8. Ln(−ei) = ln | − ei| + i.Arg(−ei) = ln e + i. −π
2 = 1 − π 2 i.
9. Ln(1 − i) = ln |1 − i| + i.Arg(1 − i) = 1 2 ln 2 + 7π 2 i.
10-16 Find all values and graph some of them in the complex plane:
10. z = ln 1 ⇒ e z = 1 ⇒ e x+iy = 1 ⇒ e x . cos y = 1 and e x . sin y = 0
⇒ y = ∓2kπ, k ∈ Z, x = 0 ⇒ ∓2kπi.
11. z = ln(−1) ⇒ e z = −1 ⇒ e x+iy = 1 ⇒ e x . cos y = −1 and e x . sin y = 0
⇒ y = ∓(2k − 1)π, k ∈ Z, x = 0 ⇒ ∓(2k − 1)πi.
12. z = ln e ⇒ e z = e ⇒ e x+iy = 1 ⇒ e x . cos y = e and e x . sin y = 0
⇒ y = ∓2kπ, k ∈ Z, e x = e ⇒ x = 1 ⇒ 1 ∓ 2kπi.
13. z = ln(−6) ⇒ e z = −6 ⇒ e x+iy = −6 ⇒ e x . cos y = −6 and e x . sin y = 0
⇒ y = ∓(2k − 1)π, k ∈ Z, e x = 6 ⇒ x = ln 6 ⇒ ln 6 ∓ (2k − 1)πi.
14. z = ln(4 + 3i) ⇒ e z = 4 + 3i ⇒ e x+iy = 4 + 3i ⇒ e x . cos y = 4 and e x . sin y = 3
⇒ y = arctan 3 4 ∓ 2nπ, n ∈ Z, e2x = 25 ⇒ e x = 5 ⇒ x = ln 5
⇒ ln 5 + (arctan 3 4 ∓ 2nπ)i.
15. z = ln(−e −i ) ⇒ e z = −e −i ⇒ e x+iy = − cos(1) + i. sin(1)
23

⇒ e x . cos y = − cos(1) and e x . sin y = sin(1).
Because suitable x and y couldn’t be found, there is no solution.
16. z = ln(e 3i ) ⇒ e z = e 3i ⇒ e x+iy = cos(3) + i. sin(3)
⇒ e x . cos y = cos(3) and e x . sin y = sin(3) ⇒ y = 3, e 2x = 1 ⇒ x = 0 ⇒ 3i.
Show that the set of values of ln(i 2 ) differs from the set of values of 2 ln i:
17. z = ln(i 2 ) ⇒ e z = −1 ⇒ e x+iy = −1 ⇒ e x . cos y = −1 and e x . sin y = 0
⇒ y = ∓(2k − 1)π, k ∈ Z, e 2x = 1 ⇒ x = 0 ⇒ ∓(2k − 1)πi ... (I).
On the other hand, z = 2 ln i ⇒ e z 2 = i ⇒ e x 2 . cos( y 2 ) = 0 and e x 2 . sin( y 2 ) = 1
Hence (I) ≠ (II).
18-21 Solve for z:
⇒ y = ∓(4k + 1)π, k ∈ Z, e x = 1 ⇒ x = 0 ⇒ ∓(4k + 1)πi ... (II).
18. ln z = (2 − 1 2 i)π ⇒ z = e(2− 1 2 i)π = e 2π .(cos(− π 2 ) + i. sin(− π 2 )) = −e2π i.
19. ln z = 0.3 + 0.7i ⇒ z = e 0.3+0.7i = e 0.3 .(cos(0.7) + i. sin(0.7)).
20. ln z = e − πi ⇒ z = e e−πi = e e .(cos(−π) + i. sin(−π)) = −e e ∼ = −15, 154.
21. ln z = 2 + π 4 i ⇒ z = e2+ π 4 i = e 2 .(cos( π 4 ) + i. sin( π 4 )) = √
2
2 e2 (1 + i).
22-28 Find the principal value of:
22. i 2i = e 2iLni = e 2i.i. π 2 = e −π ,
(2i) i = e iLn2i = e i(ln 2+ π 2 i) = e − π 2 [cos(ln 2) + i. sin(ln 2)].
23. 4 3+i = e (3+i) ln 4 = e 3 ln 4+ln 4i = e 3 ln 4 (e ln 4i ) = e 3 ln 4 (i. sin(ln 4)).
24. (1 − i) 1+i = e (1+i)Ln(1−i) = e (1+i)(ln √ 2− π 4 i) = √ 2e π 4 [cos(− π 4 + ln √ 2) + i. sin(− π 4 + ln √ 2))].
25. (1 + i) 1−i = e (1−i)Ln(1+i) = e (1−i)(ln √ 2+ π 4 i) = √ 2e π 4 [cos( π 4 − ln √ 2) + i. sin( π 4 − ln √ 2))].
26. (−1) 1−2i = e (1−2i)Ln(−1) = e (1−2i)(ln |−1|+i.Arg(−1)) = e πi+2π = e 2π (cos π + i. sin π) = −e 2π .
27. i 1/2 = e (1/2)Ln(i) = e (1/2)(ln |1|+i.Arg(i)) = e π 4 i √
= 2
2
(1 + i).
28. (3 − 4i) 1/3 = e (1/3)Ln(3−4i) = e (1/3)(ln 5+i. arctan 4 3 ) = 3√ 5[cos( 1 3 arctan 4 3 ) − i. sin( 1 3 arctan 4 3 )] ∼ =
1, 6289 − 0, 5202.i.
CHAPTER 14 - COMPLEX INTEGRATION
14.1
1-9: Parametric Representations
1. z(t) = t + i3t, 1 ≤ t ≤ 4 ⇒ x(t) = t, y(t) = 3t
• t = 1 ⇒ x(1) = 1, y(1) = 3 ⇒ starting point:(1, 3)
• t = 4 ⇒ x(4) = 4, y(4) = 12 ⇒ ending point:(4, 12)
We can sketch as follows:
24

2. z(t) = 5 − 2it, −3 ≤ t ≤ 3 ⇒ x(t) = 5, y(t) = −2t
• t = −3 ⇒ x(−3) = 5, y(−3) = 6 ⇒ starting point:(5, 6)
• t = 3 ⇒ x(3) = 5, y(3) = −6 ⇒ ending point:(5, −6)
Then we obtain figure below:
3. z(t) = z 0 + re it , 0 ≤ t ≤ 2π denotes the circle of radius r with center z 0 .
For z(t) = 4 + i + 3e it , we find z 0 = 4 + i and r = 3.
z(t) = 4 + i + 3(cos t + i sin t) = (4 + 3 cos t) + i(1 + 3 sin t) ⇒ x(t) = 4 + 3 cos t, y(t) = 1 + 3 sin t
For finding orientation, we should determine z(t) at some points.
• t = 0 ⇒ z(0) = 4 + i + 3e 0 = 7 + i
• t = π ⇒ z(π) = 4 + i + 3e iπ = 4 + i + 3(cos π + i sin π) = 4 + i + 3(−1 + i.0) = 1 + i
• t = 2π ⇒ z(2π) = 4 + i + 3e i2π = 4 + i + 3(cos 2π + i sin 2π) = 4 + i + 3(1 + i.0) = 7 + i
25

4. z(t) = 1 + i + e −πit , 0 ≤ t ≤ 2 denotes the circle of radius r = 1 with center z 0 = 1 + i.
z(t) = 1 + i + e −πit = 1 + i + cos(−πt) + i sin(−πt) = (1 + cos(πt)) + i(1 − sin(πt))
⇒ x(t) = 1 + cos(πt), y(t) = 1 − sin(πt)
For finding orientation, we should determine z(t) at some points.
• t = 0 ⇒ z(0) = 1 + i + e 0 = 2 + i
• t = 1 ⇒ z(1) = 1 + i + e −iπ = 1 + i + (cos(−π) + i sin(−π)) = 1 + i + (cos(π) − i sin(π))
= 1 + i − 1 + 0 = i
• t = 2 ⇒ z(2) = 1 + i + e −i2π = 1 + i + (cos(−2π) + i sin(−2π)) = 1 + i + (cos(2π) − i sin(2π))
= 1 + i + 1 + 0 = 2 + i
5. z(t) = e it denotes the circle of radius r = 1 with center z 0 = 0.
z(t) = e it = cos t + i sin t ⇒ x(t) = cos t, y(t) = sin t
For finding orientation and the range of graphic, we will determine z(t) at t = 0 and t = π.
• t = 0 ⇒ z(0) = e 0 = 1
• t = π ⇒ z(π) = e iπ = cos π + i sin π = −1 + 0 = −1
Hence we obtain the figure below:
6. z(t) = 3 + 4i + 5e it denotes the circle of radius r = 5 with center z 0 = 3 + 4i.
z(t) = 3 + 4i + 5e it = 3 + 4i + 5(cos t + i sin t) = (3 + 5 cos t) + i(4 + 5 sin t)
⇒ x(t) = 3 + 5 cos t, y(t) = 4 + 5 sin t, π ≤ t ≤ 2π
We will determine z(t) at t = π and z = 2π for finding orientation and range of path.
26

• t = π ⇒ z(π) = 3 + 4i + 5e iπ = 3 + 4i + 5(cos π + i sin π) = −2 + 4i
• t = 2π ⇒ z(2π) = 3 + 4i + 5e i2π = 3 + 4i + 5(cos 2π + i sin 2π) = 8 + 4i
7. z(t) = 6 cos 2t + i5 sin 2t, 0 ≤ t ≤ π. Let x(t) = 6 cos 2t and y(t) = 5 sin 2t.
• t = 0 ⇒ z(0) = 6
• t = π 2
⇒ z( π 2 ) = −6
• t = π ⇒ z(π) = 6
This parametric equation denotes the ellipse below:
8. z(t) = 1 + 2t + 8it 2 , −1 ≤ t ≤ 1 denotes parabola.
• t = −1 ⇒ z(−1) = −1 + 8i
• t = 0 ⇒ z(0) = 1
• t = 1 ⇒ z(1) = 3 + 8i
9. z(t) = 1 + 1 2 it3 , −1 ≤ t ≤ 2
• t = −1 ⇒ z(−1) = −1 − 1 2 i 27

• t = 0 ⇒ z(0) = 0
• t = 2 ⇒ z(2) = 2 + 4i
So we obtain figure below:
10. • The starting point is 1 + i. We can write as (1, 1).
• The ending point is 4 − 2i. We can write as (4, −2).
• How far does 1 have to move to get to 4 3 units. So x(t) = 1 + 3t.
• How far does 1 have to move to get to −2 −3 units. So y(t) = 1 − 3t.
We obtain z(t) = x(t) + iy(t) = (1 + 3t) + i(1 − 3t).
11. Let’s find parametric equation of the unit circle with center z 0 = x 0 + iy 0 .
z(t) = z 0 + 1.e it = x 0 + iy 0 + cos t + i sin t = (x 0 + cos t) + i(y 0 + sin t), 0 ≤ t ≤ 2π
Hence we obtain x(t) = x 0 + cos t, y(t) = y 0 + sin t, 0 ≤ t ≤ 2π.
12. Let starting point z 0 = a + ib and ending point z 1 = c + id.
28

So, we get x(t) = a + (c − a)t and y(t) = b + (d − b)t. Hence
z(t) = x(t) + iy(t) = (a + (c − a)t) + i(b + (d − b)t)
13. Let x(t) = t and y(t) = 1 t . Then we obtain parametric equation z(t) = t + i 1 t .
14. The equation of an ellipse whose major and minor axess coincide with the cartesian axis is
( x a )2 + ( y b )2 = 1
Because of y ≥ 0 we obtain following graphic:
Parametric equation is x(t) = a cos t, y(t) = b sin t, y ≥ 0.
15. At first we determine y at x = −1, x = 0 and x = 1 for helping us to get graphic.
29

• x = −1 ⇒ y = 4 − 4 = 0
• x = 0 ⇒ y = 4
• x = 1 ⇒ y = 0
Now we can sketch as follows:
Parametric equation: x(t) = t, y(t) = 4 − 4t 2 , −1 ≤ x ≤ 1
16. |z − 2 + 3i| = 4 denotes the circle of radius r = 4 with center at z 0 = 2 − 3i.
If z 0 = 2 − 3i and r = 4, then
z(t) = 2 − 3i + 4e it = 2 − 3i + 4(cos t + i sin t) = 2 − 4 cos t + i(−3 + sin t)
So we get x(t) = 2 − 4 cos t, y(t) = −3 + sin t, 0 ≤ t ≤ 2π.
17. |z + a + ib| = r denotes the circle of radius r with center at z 0 = −a − ib.
z(t) = −a − ib + re it = −a − ib + r(cos t + i sin t) = (−a + r cos t) + i(−b + r sin t)
30

Hence we get x(t) = −a + r cos t, y(t) = −b + r sin t, −2π ≤ t ≤ 0.
18. 4(x − 1) 2 + 9(y + 2) 2 = 36 denotes an ellips.
Hence we get ellips below:
4(x − 1) 2 + 9(y + 2) 2 = 36 ⇒ ( x − 1 ) 2 + ( y + 2
3 2 )2 = 1
x 0 − 1 = 0, y 0 + 2 = 0 ⇒ x 0 = 1, y 0 = −2 ⇒ z 0 = (1, −2)
Parametric equation: x(t) = 3 cos t + 1, y(t) = 2 sin t − 2, 0 ≤ y ≤ 2π
19-29 Integration:
19. Let z = x + iy. Then f(z) = Rez = x is analytic in C. So we can use the first method:
∫ z1
z 0
f(z)dz = F (z 1 ) − F (z 0 ), F (z) analytic, F ′ (z) = f(z)
Let F (z) = x2
2 . F ′ (z) = 2 x 2 = x = f(x).
∫ 1+i
20. We can sketch C as follows:
0
f(z)dz =
∫ 1+i
0
xdz = F (1 + i) − F (0) = 1 2
If z = x + iy, then f(z) = Rez = x is analytic in C. So we can use the first method.
F (z) = x2
2 is analytic in C and F ′ (z) = 2 x 2 = x = f(x).
∫ ∫ 1+i ∫ 1+i
Rez = f(z)dz = xdz = F (1 + i) − F (0) = 1 2
C
21.
31
0
0

f(z) = e 2z is analytic in C. So we can use the first method. Let F (z) = 1 2 e2z . F (z) is is analytic in C
and F ′ (z) = 1 2 2e2z = f(z). So,
∫ 2πi
πi
e 2z dz = F (2πi) − F (πi) = 1 2 e4πi − 1 2 e2πi = 1 (cos 4π + i sin 4π − cos 2π − i sin 2π) = 0
2
22. f(z) = sin z is analytic function. Hence we can use first method to find the integral.
Let F (z) = − cos z. F (z) is analytic also and F ′ (z) = sin z = f(z). So we obtain,
∫
C
sin zdz =
∫ 2i
23. At first, we sketch C:
0
sin zdz = F (2i) − F (0) = − cos 2i − (− cos 0) = − cos 2i + 1 = 1 − cosh 2
f(z) = cos 2 z =
Let F (z) =
cos 2z+1
2
is analytic in C. So, we can use first method.
sin 2z+z
2
. F (z) is analytic in C and F ′ cos 2z+1
(z) =
2
= f(z). Hence we get
∫C f(z)dz = ∫ πi
−πi cos2 zdz
= F (πi) − F (−πi)
= sin(2πi)+πi
2
− − sin(2πi)−πi
2
= sin(2πi)
= i sinh(2π)
24. We can’t use first method here because f(z) = z + z −1 isn’t defined at z = 0. Hence f(z) isn’t
analytic in C. We should use second method:
32

∫C f(z)dz = ∫ a
b f(z(t))z′ (t)dt,
z ′ = dz
dt
C : z(t) = cos t + i sin t, 0 ≤ t ≤ 2π
z ′ (t) = ie it
f(z(t)) = z(t) + 1
z(t) = eit + e −it
∫C z + z−1 dz = ∫ 2π
0 (eit + e −it ).ie it
= ∫ 2π
0
i.(e 2it + 1)dt
= i.( 1 2i e2it + t) ∣ ∣ 2π
0
= ( 1 2 e2it + it) ∣ ∣ 2π
0
= 1 2 e4iπ + 2iπ − 1 2 e0 − 0
= 1 2 (cos 4π + i sin 4π) + 2iπ − 1 2
= 1 2 + 2iπ − 1 2
= 2iπ
25. f(z) = cosh 4z is analytic in C. So, we use first method. Let F (z) = 1 4
sinh 4z. F (z) is analytic also
and F ′ (z) = 1 4
.4. cosh 4z = cosh 4z = f(z).
∫C cosh 4zdz = ∫ πi
8
cosh 4z
− πi
8
= F ( πi
8
= 1 4
) − F (−
πi
8 )
4πi
sinh(
8 ) − 1 −4πi
4
sinh(
8
)
= 1 4 (i sin π 2 + i sin π 2 )
= 1 4 .2i
= i 2
26. z(t) = t + it 2 , (−1 ≤ t ≤ 1), z ′ (t) = 1 + 2it, ¯z = t − it 2 , so that
∫ 1
−1
(t − it 2 )(1 + 2it)dt =
∫ 1
−1
(2t 3 + it 2 + t)dt = 1 3 it3 ∣ ∣∣∣
1
33
−1
= 2 3 i.

27. z(t) = ( π 4 − t) + ti, (0 ≤ t ≤ π 4 ), z′ (t) = (−1 + i), so that
∫ π
4
0
sec 2 [( π 4 − t) + ti](−1 + i)dt = ∫ πi
4
0
sec 2 udu = tan u
∣
πi
4
0
= tan πi
4 .
28. Imz 2 = 2xy is 0 on the axes. Thus the only contribution to the integral comes from the segment
from 1 to i, represented by, say,
z(t) = 1 − t + it (0 ≤ t ≤ 1).
Hence z ′ (t) = −1 + i, and the integral is
∫ 1
0
2(1 − t)t(−1 + i)dt = 2(−1 + i)
∫ 1
29. z(t) = (1 − t) + it, (0 ≤ t ≤ 1), z ′ (t) = −1 + i, so that
∫ 1
0
ze z2 2 (−1 + i)dz = (−1 + i)
∫ 1
2
0
∣ 1
e u du = (−1 + i) eu ∣∣∣ 2
u ′
0
(t − t 2 )dt = 1 (−1 + i).
3
0
= (−1 + i) e( 1 2
= (1 − i)(e −1
2 i + e 1 2 ).
− t + i(1 − t)t)
(1 − t) + it
∣
1
0
14.2
1,2,6 :
1.
f(z) = Rez is defined and differentiable at all points of D.
=⇒ f(z) is analytic. Then by Cauchy’s integral theorem ∫ Rezdz = 0.
2.
f(z) = 1
3z−Πi and let C be the unit circle. 3z − πi = 0 =⇒ z = π 3 i.
Since this point lies outside C, f is defined and differentiable on C and inside C.
34

=⇒ f is analytic. Then by Cauchy’s integral theorem ∫ dz
3z−π = 0.
6.
f(z) = sec( z 1
2
) =⇒ f(z) =
cos( z 2 )
cos( z 2
) = 0 =⇒ z = (2k + 1)π, k ∈ Z
But these points lie outside the unit circle. Then f is defined and differentiable on C and inside C.
=⇒ f is analytic. Then by Cauchy’s integral theorem ∫ sec( z 2
)dz = 0.
12-17 :
12.
a) f(z) = 1
z 2 +4
z 2 + 4 = 0 =⇒ z 2 = −4 =⇒ z = ±2i.
These points lie outside the region.
=⇒ f is analytic.
=⇒ ∫ dz
z 2 +4 = 0.
b) f(z) = 1
z 2 +4
35

z 2 + 4 = 0 =⇒ z 2 = −4 =⇒ z = ±2i.
These points lie inside the region. =⇒ f isn’t analytic in the region.
13.
Since z 2 is differentiable in the region, it is analytic. By Cauchy’s integral theorem ∫ z 2 dz = 0.
14.
a) f(z) = 1 z
From here if z = 0, f isn’t analytic.
=⇒ When z ≠ 0, f is analytic.
İf z ≠ 0, İntegral of f is 0.
b) f(z) = cos z
z 6 −z 2
z 6 − z 2 = 0 =⇒ z 2 (z 4 − 1) = 0 =⇒ z = 0, z = ±i, z = ±1.
=⇒ When z ≠ 0, z ≠ ±i and z ≠ ±1, f is analytic.
İf z ≠ 0, z ≠ ±i and z ≠ ±1, İntegral of f is 0.
c) f(z) = e 1 z
z 2 +9
z 2 + 9 = 0 =⇒ z 2 = −9 =⇒ z = ±3i
From e 1 z , z = 0.
=⇒ When z ≠ 0 and z ≠ ±3i, f is analytic.
İf z ≠ 0 and z ≠ ±3i, İntegral of f is 0.
36

15. We shall remember Example 4:
∫
C
dz
z 2 = 0
where C is the unit circle. This result does not follow from Cauchy’s theorem, because f(z) = 1
z 2
not analytic at z = 0. Hence the condition that f be analytic in D is sufficient rather than necessary for
Cauchy’s integral theorem to be true.
Then they can be deformed each other. So integral of f is 0.
16. İf C is the unit circle, ∫ f(z)dz = 3
İf C is |z| = 2, ∫ f(z)dz = 5.
is
f isn’t analytic in the annulus 1 < |z| < 2 because of the principle of deformation of path.
17.
a) C 1 : x : 0 → π and y : 0 → π
y = x =⇒ dy = dx.
İf z = x + iy, dz = dx + idy.
=⇒ ∫ C 1
cos zdz = ∫ cos(x + iy)(dx + idy) = ∫ π
0 cos((1 + i)x)(1 + i)dx = sin((1 + i)x)∣ ∣ π 0 =
sin((1 + i)π) − sin 0 = sin(π + iπ) = − sin(iπ).
b) ∫ C cos zdz = ∫ C 2
+ ∫ C 3
C 2 : x : 0 → π and y = 0 =⇒ dy = 0
İf z = x + iy, dz = dx + idy = dx.
=⇒ ∫ C cos zdz = ∫ π
0 cos(x + iy)dx = ∫ π
0 cos xdx = sin x∣ ∣ π 0 = sin π = 0
C 3 : x = π =⇒ dx = 0 and y : 0 → π
37

İf z = x + iy, dz = dx + idy = idy.
=⇒ ∫ C 3
cos zdz = ∫ π
0 cos(x + iy)idy = ∫ π
0 i cos(π + iy)dy = sin(π + iy)∣ ∣ π 0
sin π = − sin(iπ).
=⇒ ∫ C
cos zdz = 0 − sin(iπ) = − sin(iπ).
Then integral of cos z is independent of path.
= sin(π + iπ) −
19-21 :
19.
2z − i = 0 =⇒ z = i 2
lies inside the region.
Then let be |2z − i| = ε.
=⇒ 2z − i = εe iθ
=⇒ z = 1 2 (i + εeiθ )
=⇒ dz = 1 2 (iεeiθ )dθ
=⇒ ∫ dz
2z−i = ∫ 2π
0
20.
1
2 iεeiθ dθ
= 1 εe iθ 2i2π = πi.
∫
C tanh zdz = ∫ C
sinh z
cosh z dz
cosh z = 0 =⇒ z = ± πi
2 , ± 3πi
2 , ...
All these points lie outside C.
=⇒ tanh z is analytic in the region.
=⇒ ∫ C tanh zdz = 0. 38

21.
İf z = x + iy, then f(z) = Re2z = 2x is analytic.
=⇒ ∫ C
Re2zdz = 0.
14.3
1.We should use Cauchy’s integral formula to solve this question.
We can sketch C : |z − i| = 2 as follows:
∮
C
f(z)
z−z 0
dz = 2πif(z 0 )
Let g(z) = z2 −4
z 2 +4 .We know that z2 + 4 = (z − 2i)(z + 2i).
z 0 − 2i = 0 ⇒ z 0 = 2i
z 1 + 2i = 0 ⇒ z 1 = −2i
C encloses the point z 0 = 2i but don’t encloses the point z 1 = −2i, where g(z) is not analytic. Let
f(z) = z2 −4
z+2i
, D be the union of C and the interior part of C. f(z) is analytic in D. So we can use the
formula.
∮
z 2 −4
C z 2 +4 dz = ∮ z 2 −4
C z+2i . 1
z−2i dz = 2πi.f(2i) = 2πi. (2i)2 −4
2i+2i
= 2πi. 4i2 −4
4i
= 2πi. −8
4i
== −4π
4. We sketch C : |z| = π/2 as follows:
39

Let D be the union of C and the interior part of C. π/2 = 1, 57079633 , −2i, 2i /∈ D and therefore
f(z) = z2 −4
is analytic in D. By using Cauchy’s integral theorem, we obtain
z 2 +4
∮
z 2 − 4
z 2 + 4 dz = 0
5-17: Contour Integration
C
5. C : |z − 1| = 2
Let g(z) = z+2
z−2 . If z − 2 = 0, then the singular point z 0 = 2 is enclosed by C. Let D be the union of C
and the interior part of C. f(z) = z + 2 is analytic in D. Hence, by use of Cauchy’s integral formula, we
obtain
6. C : |z| = 1
∮
C
z + 2
dz = 2πi.f(2) = 2πi.4 = 8πi
z − 2
g(z) =
e3z 1
3z − i = 3 e3z
z − i 3
Let D be the union of C and the interior part of C.
z − i 3 ⇒ z 0 = i 3 ∈ D
f(z) = 1 3 e3z is analytic in D. Hence by use of Cauchy integral formula, we get
∮ ∮ 1
3
g(z)dz =
e3z
C
C z − i dz = 2πi.f( i 3 ) = 2πi.1 3 e3 i 2
3 =
3 πiei
3
7. C : |z| = 1
40

g(z) =
sinh πz
z 2 − 3z
=
sinh πz
z(z − 3)
Singular points of g are z 0 = 0 and z 1 = 3. Let D be the union of C and the interior part of C. Then
z 1 = 3 /∈ D but z 0 = 0 ∈ D. f(z) =
we get
12.
∮
C
∮
g(z)dz =
C
sinh πz
z−3
is analytic in D. Hence by use of Cauchy integral formula,
sinh πz
z − 3 .1 0
dz = 2πi.f(0) = 2πisinh
z −3
= 2π sin(i.0) = 0
−3
Let g(z) = tan z
z−i . Singular point z 0 = i is enclosed by C. f(z) = tan z is analytic in D when D is the
union of C and the interior part of C. So by the use of Cauchy integral formula, we get
∮
tan z
dz = 2πi tan i
z − i
13.
C
g(z) = e−3πz 1
2z + i = 2 e−3πz
z + i 2
Singular point z 0 = − i 2 is enclosed by C. f(z) = 1 2 e−3πz is analytic in D when D is the union of C and
the interior part of C. So by use of Cauchy integral formula, we get
∮
e −3πz
1
2z + i
∮C
dz = dz = 2πi. 1 2 e− 3 2 πi =
2 e−3πz
C
z + i 2
15. We sketch C : |z − 4| = 2 as follows:
41
πi
3
2 (cos π + i sin π) = −2 3 πi

Let g(z) = ln(z−1)
z−5
. Singular point z 0 = 5 is enclosed by C. Let D be the union of C and the interior
part of C, f(z) = ln(z − 1). Then f(z) is analytic in D. Therefore, by use of Cauchy integral formula,
we get
∮
C
ln(z − 1)
dz = 2πif(5) = 2πi ln 4
z − 5
16. Let C 1 : |z| = 3 and C 2 : |z| = 1. We sketch C = C 1 ∪ C 2 = 2 as follows:
g(z) =
sin z
z 2 − 2iz =
sin z
z(z − 2i)
Singular points of g are 0 and 2i. Only z 0 = 2i is contained in the ring-shaped domain bounded by C 1
and C 2 . f(z) = sin z
z
is analytic on that domain. Hence by use of Cauchy integral formula, we get
∫
sin z
2i
z 2 dz = 2πi.f(2i) = 2πi.sin = π sin 2i
− 2iz 2i
18.Let C a simple closed path enclosing z 1 and z 2 .
C
By use of Cauchy integral formula, we obtain
∮
C (z − z 1) −1 (z − z 2 ) −1 dz = ∮ 1
C (z−z 1 )(z−z 2 ) dz = ∮ 1 1
C 1 z−z 2
.
z−z 1
dz + ∮ 1 1
C 2 z−z 1
.
z−z 2
dz
= 2πi[
1
z−z 2
] z=z1 + 2πi[
1
z−z 1
] z=z2 = 2πi
1
z 1 −z 2
+ 2πi
1
z 2 −z 1
= 2πi(
1
z 1 −z 2
− 1
z 1 −z 2
) = 0
14.4
1-8: Contour Integration Let C : |z| = 2 and D be the union of C and the interior part of C.
42

1. We use Theorem 1 on the page 658.
f(z) = cosh 3z is analytic in D.
f n (z 0 ) = n! ∮
2πi C
Because n + 1 = 5, we need to know f (4) (z):
z 5 0 = 0 ⇒ z 0 = 0 ∈ D
f(z)
(z − z 0 )
∮C
n+1 ⇒ f(z)
(z − z 0 ) n = f n (z 0 ).2πi
, n = 1, 2, ...
n!
f (1) (z) = 3 sinh 3z, f (2) (z) = 9 cosh 3z, f (3) (z) = 27 sinh 3z, f (4) (z) = 81 cosh 3z
Hence, we get
∮
C
cosh 3z
z 5 dz = f 4 (0).2πi 81 cosh 0.2πi
= = 81.1.2πi
4! 4.5.2.1 4.3.2.1 = 27πi
4
2. z 0 = πi
2 ∈ D. f(z) = sin z is analytic in D. We need to know f (3) (z):
f (1) (z) = cos z, f (2) (z) = − sin z, f (3) (z) = − cos z
Hence, we obtain
∮
C
sin z
(z − πi dz = f (3) ( πi
2 ).2πi
2 )4 3!
=
πi
− cos(
2 ).2πi =
3!
πi
− cos(
2 ).πi = − cosh( π 2 ).πi
3
3
3. z 0 = πi
2 ∈ D and f(z) = ez cos z is analytic in D.
f(z) = e z cos z ⇒ f (1) (z) = e z cos z − e z sin z = e z (cos z − sin z)
So,
∮
C
e z cos z
(z − π dz = f (1) ( π 2 ).2πi = e π π
2 (cos
2 )2 1!
2 − sin π 2 ).2πi = −e π 2 .2πi
4. z 0 = 0 ∈ D and f(z) = cos z is analytic in D.
f (1) (z) = − sin z, f (2) (z) = − cos z, f (3) (z) = sin z,
f (4) (z) = cos z, f (5) (z) = − sin z, f (6) (z) = − cos z,
· · ·
43

Hence we see that
when k ∈ {0, 1, 2, ...}.
∮
n = 2 2k ⇒
C
∮
n = 2 2k−1 ⇒
n = 2 2k ⇒ f (2n) (z) = − cos z
n = 2 2k−1 ⇒ f (2n) (z) = cos z
cos z
z 2n+1 dz = f (2n) (0).2πi − cos 0.2πi
= = −2πi
2n!
2n! (2n)!
C
cos z
z 2n+1 dz = f (2n) (0).2πi cos 0.2πi
= = 2πi
2n! 2n! (2n)!
9.Let C : |z| = 1 and D be the union of C and the interior part of C.
(1 + 2z) cos z
(2z − 1) 2 =
f(z) = 1 4
.(1 + 2z) cos z is analytic in D.
1
4
.(1 + 2z) cos z
(z − 1 2 )2
f (1) (z) =
2. cos z + (1 + 2z).(− sin z)
4
=
2 cos z − (1 + 2z) sin z
4
Hence, we get
∮
C
(1+2z) cos z
dz = ∮ (2z+1) 2 C
1
4
.(1+2z) cos z
dz = f (1) ( 1
(z− 1 2 ).2πi
2 )2
1!
= ( 2 cos 1 2 −2 sin 2
1 ).2πi
4
1
= πi(cos 1 2 − sin 1 2 )
10. Let C 1 : |z| = 5 (counterclockwise), C 2 : |z − 3| = 3 2 (clockwise) and C = C 1 ∪ C 2
Let g(z) =
sin 4z
. z
(z−4) 3 0 = 4 isn’t contained the ring-shaped domain bounded by C 1 and C 2 . By use of
Cauch’s integral theorem we get
∫
C
sin 4z
(z − 4) 3 dz = 0
11. We can the sketch ellipse C : 16x 2 + y 2 = 1 as follows:
44

z 0 = 0 ∈ D. If we take f(z) = tan πz then f(z) is analytic in D and f (1) (z) = 1 π sec2 πz.
∮
tan πz
z 2 dz = 1 π sec2 0.2πi = 1
cos 2 .2i = 2i
0
15.1
1-3 Sequences :
C
CHAPTER 15 - POWER SERIES, TAYLOR SERIES
1. Let z n = (−1) n + i
2 n . z n is bounded because
As
and by ratio test, we get z n is convergent.
2. Let z n = e −nπi/4 . z n is bounded since
Because
|z n | = |(−1) n + i
√
2 n | = (−1) 2n + 1
√
2 2n = 1 + 1
√
lim
n→∞
1 + 1 = 1.
22n √
∣ z n+1
∣ |z n+1 | 1 + 1
2
= = √
2n+2
< 1
z n |z n |
1 + 1
2 2n
|z n | = |e −nπi/4 | =
√
cos 2 ( nπ 4 ) + sin2 ( nπ 4 ) = 1.
2 2n
∣ z n+1
∣ ∣ = e −(n+1)πi/4
√
∣ = |e −πi/4
z n e −nπi/4 | = cos 2 ( π 4 ) + sin2 ( π 4 ) = 1
and by ratio test, we have z n is convergent.
3. Let z n = (−1)n
n+i . z n is bounded as
|z n | = | (−1)n
n + i | = 1
|n + i| = 1
√
n 2 + 1
45

Because
lim
n→∞
1
√
n 2 + 1 = 0.
∣ z √
n+1
∣ ∣ = n + i
∣
|n + i| n =
z n n + 1 + i |n + 1 + i| = 2 + 1
√
(n + 1) 2 + 1 < 1
and from ratio test, we have z n is convergent.
16-18 Series :
16. Let
∞∑ (10 − 15i) n
n=0
n→∞
n!
lim ∣ z n+1
∣ = lim ∣
z n n→∞
and by ratio test we get this result.
17. Let
and z n = (−1)n (1+2i) 2n+1
(2n+1)!
. This serie is convergent from ratio test because
18. Let
∞∑ (−1) n (1 + 2i) 2n+1
n=0
∞∑
n=0
n→∞
and we know that
15.2
(2n + 1)!
and z n = (10−15i)n
n!
. This serie is convergent because
(10−15i) n+1
(n+1)!
(10−15i) n
n!
∣ = lim
n→∞
∣
√
10 − 15i
∣ = lim
n + 1 n→∞
325
(n + 1) 2 = 0
lim ∣ z n+1
∣ ∣
(1 + 2i)
= lim ∣(−1)
2 ∣
z n n→∞ (2n + 3)(2n + 2)
i n
n 2 − 2i and z n =
∞∑
n=0
3-5 Radius of Convergence :
∣ = lim
n→∞
5
(2n + 3)(2n + 2) = 0.
in . This serie is convergent, we now explain this situation :
n 2 −2i
∣
i n
n 2 − 2i | = |in |
|n 2 − 2i| = 1
|n 2 − 2i| ≤ 1 n 2
1
is convergent. Then via comparison test, we can say that this serie is convergent.
n2 ∞∑ (z + i) n
3. Let
. First, we determine the center.
n=1
n 2
z + i = 0 ⇒ z = −i
is the center point. On the other hand, z n = (z+i)n
n 2
and because
lim
n→∞
radius of convergence is 1.
∣
1
n 2
1
n→∞
(n+1) 2 | = lim
∣
(n + 1)2 ∣ n 2 + 2n + 1
= lim
n 2 n→∞ n 2 = 1,
46

∞∑
4. Let
n=0
n n
n! (z + 2i)n . First of all, center is −2i since
Let z n = nn
n! (z + 2i)n .As
lim
n→∞
∣
n n
n!
(n+1) ( n+1)
(n+1)!
radius of convergence is 1 e .
| = lim
n→∞
z + 2i = 0 ⇒ z = −2i.
n n (n + 1)
= lim
(n + 1) n+1 n→∞
n n
(n + 1) n = lim
( n ) n 1 =
n→∞ n + 1 e ,
5. Let
∞∑
n=0
n!
n n (z + 1)n . The center point is −1 since
Let z n = n!
n n (z + 1) n . Then
lim
n→∞
we say that radius of convergence is e.
∣
z + 1 = 0 ⇒ z = −1.
n!
n n
(n+1)!
(n+1) ( n+1)
| = lim
n→∞
(n + 1) n = e,
n
15.3
1-3 Radius of convergence by differentiation or integration :
1. Let
∞∑
n=2
n(n − 1)
3 n (z − 2i) n . Since
lim
n→∞
we say that radius of convergence is 1 4 .
∣
4 n
n(n+1)
4 n+1
(n+1)(n+2)
n + 2
| = lim
n→∞ 4n = 1 4 ,
2. Let
∞∑
n=1
4 n
n(n + 1) zn . Because
radius of convergence is found that 3.
3. Let
∞∑
n=1
lim ∣ n(n−1)
3 n
n→∞
| = lim
n(n+1) n→∞
3 n+1
3n − 3
n + 1 = 3,
n
2 n (z + i)2n . Radius of convergence for this serie is 2 in that
lim
n→∞
∣
n
2 n
n+1
n→∞
2 n+1 | = lim
47
2n
n + 1 = 2.

15.4
1-3 Taylor and Maclaurin series :
1. Let center be 0 for e −2z . We know that
e z =
∞∑
n=0
z n
n! = 1 + z + z2
2! + . . .
at z = 0. But here the function is e −2z , then Maclaurin serie for this function is
∞∑
e −2z (−2z) n
=
n!
n=0
Radius of convergence for this serie is ∞ in that
= 1 + (−2z) + 4z2
2!
+ . . .
lim R = lim
n→∞ n→∞
∣
∣ (−2)n
n!
(−2) n+1
(n+1)!
n=0
−(n + 1)
| = lim = ∞.
n→∞ 2
1
2. Let center be 0 for
(1−z 3 ) . 1
∞
1 − z = ∑
z n = 1 + z + z 2 + z 3 + . . .
is known at 0. Then Maclaurin serie for
1
(1−z 3 ) is
1
∞
1 − z 3 = ∑
z 3n = 1 + z 3 + z 6 + z 9 + . . .
n=0
Radius of convergence for this serie is 1 since
3. Let center be −2i for e z .We know that
e z =
∞∑
n=0
lim R = 1
n→∞
z n
n! = 1 + z + z2
2! + . . .
at z = 0. But here the center is −2i, then Taylor serie for this function is
e z =
∞∑
n=0
z n
n!
Radius of convergence for this serie is ∞ in that
= 1 + (z + 2i) +
(z + 2i)2
2!
+ . . .
lim
n→∞
∣
1
n!
1
(n+1)!
| = lim
n→∞ n + 1 = ∞.
15.5
48

1-3 Uniform Convergence :
1. Let
∞∑
(z − 2i) 2n and |z − 2i| ≤ 0.999. We know that
n=0
1
1 − z = 1 + z + z2 + z 3 + . . .
1
∞
1 − (z − 2i) 2 = ∑
(z − 2i) 2n = 1 + (z − 2i) 2 + (z − 2i) 4 + (z − 2i) 6 + . . .
n=0
Because |z − 2i| ≤ 0.999, this serie converges uniformly.
2. Let
∞∑
n=0
z 2n+1
(2n + 1)! and |z| ≤ 1010 .999. Since
sinh z =
∞∑
n=0
z 2n+1
(2n + 1)! = z + z3
3! + z5
5! + . . .
this Maclaurin series of sinh z converges uniformly on every bounded set.
3. Let
∞∑
n=0
π n
n 4 z2n and |z| ≤ 0.56. From ratio test, we have
lim L = lim
n→∞ n→∞
∣
π n+1
(n+1) 4
π n
n 4
∣ ∣ = lim
n→∞
and |z| ≤ 0.56, this serie converges uniformly on everywhere.
πn 4
(n + 1) 4 = π > 1,
9-11 Power series :
9. Let
∞∑ (z + 1 − 2i) n
4 n . By ratio test, we get
n=0
lim L = lim
n→∞ n→∞
∣
1
4 ( n+1)
1
4 n ∣ ∣ = lim
n→∞
then we can say that this serie converges uniformly on everywhere.
10. Let
∞∑
n=0
(z − i) 2n
. Since
(2n)!
cosh z =
∞∑
n=0
1
4 = 1 4 < 1,
z 2n
(2n)! = 1 + z2
2! + z4
4! + . . .
this Taylor series of cosh z with center i converges uniformly on every bounded set.
49

∞∑
11. Let
n=1
(−1) n
2 n n zn . By ratio test, we get
lim L = lim
n→∞ n→∞
∣
(−1) n+1
2 n+1 (n+1)
(−1) n
2 n n
∣ = lim
n→∞
−2n − 2
n
then we can say that this serie converges uniformly on everywhere.
= −2 < 1,
50