COMPLEX ANALYSIS

Math 252 Complex Analysis Page 1

Exercises1:

Question1. Prove that if number is equal to its conjugate, it is necessarily real. Proof:

Let where , z a ib a b

Assume that . We need to show that is real.

Since ,

a ib a ib

Question2. Prove that any number is equal to the conjugate of its conjugate. Proof:

Let be any number , where , z a ib a b

We need to show that . We need to show that is real.

Since ,

z a ib z

Thus,

Question3. Prove that if both are real, then either are both real or Proof:

Let 1 1 1 2 2 2 1 2 1 2 and be any two numbers , where , , , z a ib z a ib a a b b

Assume that are real s.t. Since

z z z

z z

2 0i b

0b

z z z

z a ib

z z

1 2 1 2 and z z z z

1 2 1 1 2 2 1 2 1 2 3 z z a ib a ib a a i b b a

1 2 1 2 and z z z z 1 2 and z z 1 2= z z

1 2 1 1 2 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 2 1 4 z z a ib a ib a a ia b ia b bb a a bb i a b a b a

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 2

1 2 3 1 2 and 0 a a a b b ,

,

,

If then are real.

If and then 1 1 1 2 2 2= =z z a ib a ib

Question4. Prove that if both if the product of two complex numbers is zero, at least one of the numbers must be zero. Proof:

Let 1 1 1 2 2 2 1 2 1 2 and be any two numbers , where , , , z a ib z a ib a a b b

,

Substitute into the 1 2 2 1 0 a b a b

1 22 2 1

2

0 b b

b a ba , multiply by both sides, then

2 2 2 2

1 2 2 1 1 2 2 0 bb a b b b a , 1 0b , then that is 1 0z

Or 22 2

2 2 2 0 b a z , 2 2 20 , 0 z a b

Thus, at least one of the numbers is zero 1 20 and 0 z z .

1 2 1 2 4 1 2 2 1 and 0 a a bb a a b a b

1 2 2 2 2 1 2 0 a b a b b a a

1 2 and zz2 10 also 0 b b

2 1 20 or b a a

1 2a a

1 2 b b

1 2 b b

1 2 1 1 2 2 1 2 1 2 2 1 1 2 1 2 1 2 1 2 2 1 0 z z a ib a ib a a ia b ia b bb a a bb i a b a b

1 2 1 2 0 a a bb 1 21

2

b b

aa

2a

1 21

2

0 b b

aa

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 3

Question5. Prove that 1 1

2 2

.

z z

z z

Proof:

Let 1 1 1 2 2 2 1 2 1 2 and be any two numbers , where , , , z a ib z a ib a a b b

1 2 1 2 1 2 2 1 1 2 2 11 1 1 1 2 2 1 2 1 2

2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2

1 2 2 11 2 1 2

2 2 2 2

2 2 2 2

1

=

a a b b i a b a b a b a ba ib a ib a ib a a b bi

a ib a ib a ib a b a b a b

a b a ba a b bi

a b a b

a a

2 1 2 1 2 2 1

2 2

2 2

2 1 1 2 1 1

2 2 2 2

2 2 1 1

2 2 2 2

1 1 1

2 2 2

=

=

=

b b ia b ia b

a b

a a ib ib a ib

a ib a ib

a ib a ib

a ib a ib

a ib z

a ib z

Question6. Verify that satisfies the equation Proof:

Substitute into the equation

1 3

2

iz 2 1 0. z z

1 3

2

iz 2 1 0. z z

2

21 3 1 3 1 3 1 3 1 3

1 1 12 2 2 2 2

1 2 3 3 1 3 = 1

4 2

1 3 1 3 = 1 0 , it satisfies this equation.

2 2 2 2

i i i i iz z

i i

i i

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 4

Question7. Reduce each of the following expression to the form a.

b.

Question8. Solve the equation . Solution: Let where , then Thus, . Question9. Find all the distinct fourth roots of -1. Solution: First of all, we have to write z in polar form.

The polar form of the complex number

2 2

1 2 i i

. z a ib

2 2

1 2 1 1 2 2 1 2 1 4 4 1 3 2 i i i i i i i i i

1 1

1 1

i i

i i

1 1 1 11 1 1 2 1 1 2 1 42

1 1 1 1 1 1 2

i i i ii i i i ii

i i i i

0 zz z z

z a ib , a b

2 2 2 2 2 0 zz z z a ib a ib a ib a ib a b a ib a ib a b bi

2 2 0 and 2 0, 0 a b b b 0a

0z

2 21, 1, 0, 1 cos sin , , 1 0 1 z a b z i r z

cos 2 sin 2 cos 2 sin 2 , 0, 1, 2,... z r k ir k r k i k k

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 5

We find that the four fourth roots of -1 are given by the expression.

is nth roots of z.

or explicitly

for

for

for

for

Question10. Find all the distinct cube roots of i. Solution: First of all, we have to write z in polar form.

The polar form of the complex number

1 12 2

cos sin , 0, 1, 2,...

n nk

k kr z r i k

n n

1 1

4 42 1 2 1

cos sin , 0,1,2,34 4

k

k kr z r i k

0k 1

1 1 1cos sin 1

4 4 2 2 2

r i i i

1k 2

3 3 1 1 1cos sin 1

4 4 2 2 2

r i i i

2k 3

5 5 1 1 1cos sin 1

4 4 2 2 2

r i i i

3k 4

7 7 1 1 1cos sin 1

4 4 2 2 2

r i i i

2 2, 0, , cos 2 sin 2 , , 0

2 2 2

z i a b i z i k i k r z i i

cos 2 sin 2 cos 2 sin 2 , 0, 1, 2,... z r k ir k r k i k k

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 6

We find that the four fourth roots of i are given by the expression.

is nth roots of z.

or explicitly

for

for

for

Question11. Express the complex number in polar form and find its distinct fourth roots. Solution: First of all, we know that

The polar form of the complex number

1 12 2

cos sin , 0, 1, 2,...

n nk

k kr z r i k

n n

1 1

3 3

2 22 2cos sin , 0,1,2

3 3

k

k k

r z i i k

0k 1

3 1 1cos sin 3

6 6 2 2 2

r i i i

1k 2

5 5cos sin

6 6

r i

2k 3

9 9cos sin

6 6

r i

1 1

22

8 3 58 8 3 , 8, 8 3, tan tan 3 60 300 ,

8 3

8 8 3 16

z i a b

r z

cos 2 sin 2 cos 2 sin 2 z r k ir k r k i k

8 8 3 i

1tan or tan , tan

y b b

x a a

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 7

We find that the four fourth roots of i are given by the expression.

is nth roots of z.

or explicitly

for

for

for

for

1 12 2

cos sin , 0, 1, 2,...

n nk

k kr z r i k

n n

1 1

4 4

5 52 2

3 316 cos sin , 0,1,2,34 4

k

k k

r z i k

0k 1

5 52 cos sin

12 12

r i

1k 2

11 112 cos sin

12 12

r i

2k 3

17 172 cos sin

12 12

r i

3k 4

23 232 cos sin

12 12

r i

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 8

Question12. Show that Solution:

Let 1 1 1 2 2 2 1 2 1 2 and be any two numbers , where , , , z a ib z a ib a a b b

Let , substitute all of them into the

2 2 2 2

1 2 1 2 1 22 2 z z z z z z

2 2 2 2

1 2 1 2 1 1 2 2 1 1 2 2

2 2

1 2 1 2 1 2 1 2

2 2 2 2

1 2 1 2 1 2 1 2

2

1 1 2

=

= 2

z z z z a ib a ib a ib a ib

a a i b b a a i b b

a a b b a a b b

a a a

2 2 2 2 2 2 2

2 1 1 2 2 1 1 2 2 1 1 2 2

2 2 2 2

1 1 2 2

2 2

1 2

2 2 2

= 2 2

= 2 2

a b b b b a a a a b b b b

a b a b

z z

2 22 2 2 2

1 1 1 2 2 2 and z a b z a b2 2

1 2 1 2 z z z z