complex analysis 2

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COMPLEX ANALYSIS Math 252 Complex Analysis Page 1 Exercises2: Triangle Inequality, Roots of the Complex Numbers Question1. a. Verify that 1 1 3,1 3, 1 , 2,1 5 10 Solution: Since 1 2 1 1 2 2 1 2 1 2 1 2 2 1 , , , zz x y x y xx yy xy xy b. 5 1 2 3 2 i i i i Solution: 5 5 5 5 5 1 2 3 2 3 1 3 1 3 3 3 10 3 10 5 5 = 10 10 2 i i i i i i i i i i i i ii c. 4 1 4 i Solution: 4 1 1 1 1 1 1 2 1 1 2 1 2 2 4 i i i i i i i i i 1 1 1 1 1 1 3,1 3, 1 , 3 3 1 1,3 1 1 3 , 10, 0 , 5 10 5 10 5 10 1 1 1 1 = 10 0 , 10 0 2,1 5 10 10 5

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Complex Analysis 2

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COMPLEX ANALYSIS

Math 252 Complex Analysis Page 1

Exercises2: Triangle Inequality, Roots of the Complex

Numbers

Question1.

a. Verify that 1 1

3,1 3, 1 , 2,15 10

Solution:

Since 1 2 1 1 2 2 1 2 1 2 1 2 2 1, , , z z x y x y x x y y x y x y

b.

5

1 2 3 2

i

i i i

Solution:

5 5 5 5 5

1 2 3 2 3 1 3 1 3 3 3 10 3 10

5 5 =

10 10 2

i i i i i i i i i

i i i

i i

c. 4

1 4 i

Solution:

4

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

1 1 1 1 1 13,1 3, 1 , 3 3 1 1 , 3 1 1 3 , 10,0 ,

5 10 5 10 5 10

1 1 1 1 = 10 0 , 10 0 2,1

5 10 10 5

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 2

Question2. Show that 2 21 1 2 z z z

Solution: Since and then

Let where , z a ib a b

Thus, 2 21 1 2 z z z

Question3. Verify that each of the two numbers satisfies the equation Solution:

For , we have For , we have

Thus, satisfies the equation Question4. Show that a. Solution:

Let where , z a ib a b

, and Thus,

2 2Re 1 Re 1 2 z z z

2 2Im 1 Im 1 2 z z z

2 21 1 2 , z z z z

2 2 2 2 2 21 1 1 2 1 1 2 2 2 z a ib a a bi b a a b i b ab

22 2 2 2 21 2 1 2 1 2 2 2 1 2 2 2 z z a ib a ib a i b a i ab b a a b i b ab

1 z i2 2 2 0 z z

1 z i 22 2 2 1 2 1 2 1 2 1 2 2 2 0 z z i i i i

1 z i 22 2 2 1 2 1 2 1 2 1 2 2 2 0 z z i i i i

1 z i 2 2 2 0 z z

Im Reiz z

Im iz a iz i a ib ia b Re z a

Im Reiz z

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 3

b. Solution:

Let where , z a ib a b

, and Thus, c. Solution:

Let where x, z x iy y

Since , and then

Re iz b iz i a ib ia b Im z b

Re Im iz z

Re Im iz z

1, 0

1 z z

z

11 zz

1

2 2 2 2,

x yz

x y x y

1 11

1

1 2 2 2 2 2 2

2 2

2 2 2 2

1 1 1, ,

1

, ,

x yz x y

z x y x y x y

z

x yx y x y z

x y x y

COMPLEX ANALYSIS

Math 252 Complex Analysis Page 4

Question5. Solve the equation for by writing and then solving a pair of simultaneous equations in x and y. Solution: Let , then implies that

2 2 1 0 x y x and 1

2 0, 2 1 0, 2

xy y y x x ,

if 0y then 2 1 0 x x that is x is complex but we found x is real. Substitute

1

2 x into the

2 2 1 0 x y x

2

2 2 21 11 1 0

2 2

x y x y ,

3

2 y .

Thus, 1 3

, ,2 2

z x y

2 1 0 z z ,z x y

, , , 1,0 0,0 x y x y x y

,z x y 2 1 0 z z

2 2

2 2

2 2

, , , 1,0 0,0

, , 1,0 0,0

,2 , 1,0 0,0

1,2 0,0

x y x y x y

x y xy xy x y

x y xy x y

x y x xy y