# book - Exercise Solutions for Complex Analysis Stephen...

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Unformatted text preview: Exercise Solutions for Complex Analysis Stephen Taylor April 1, 2005 2 Chapter 1 The Complex Number System 1.1 Pages 2-3 1. Find the real and imaginary parts for each of the following: In the following let the complex variable z = x + iy for { x,y } ∈ R (i) 1 z = 1 x + iy = 1 x + iy ( x- iy ) ( x- iy ) = x- iy x 2 + y 2 ∴ Re( z ) = x x 2 + y 2 and Im( z ) =- y x 2 + y 2 (ii) Let a ∈ R z- a z + a = x + iy- a x + iy + a ( x- iy + a ) ( x- iy + a ) = x 2 + y 2- a 2 + 2 ayi x 2 + y 2 + 2 ax + a 2 ∴ Re( z ) = x 2 + y 2- a 2 x 2 + y 2 + 2 ax + a 2 and Im( z ) = 2 ay x 2 + y 2 + 2 ax + a 2 (iii) z 3 = ( x + iy ) 3 = x 3- 3 xy 3 + i 3 x 2 y- iy 3 ∴ Re( z ) = x 3- 3 xy 3 and Im( z ) = 3 x 2 y- y 3 (iv) 3 4 CHAPTER 1. THE COMPLEX NUMBER SYSTEM z = 3 + 5 i 1 + 7 i = 3 + 5 i 1 + 7 i 1- 7 i 1- 7 i = 19- 8 i 25 ∴ Re( z ) = 19 / 25 and Im( z ) =- 8 / 25 (v) z =- 1 + i √ 3 2 3 = 1 ∴ Re( z ) = 1 and Im( z ) = 0 (vi) z =- 1- i √ 3 2 6 = 1 ∴ Re( z ) = 1 and Im( z ) = 0 (vii) i n for 2 ≤ n ≤ 8 n = 2 : Re( z ) =- 1 and Im( z ) = 0 n = 3 : Re( z ) = 0 and Im( z ) =- 1 n = 4 : Re( z ) = 1 and Im( z ) = 0 n = 5 : Re( z ) = 0 and Im( z ) = 1 n = 6 : Re( z ) =- 1 and Im( z ) = 0 n = 7 : Re( z ) = 0 and Im( z ) =- 1 n = 8 : Re( z ) = 1 and Im( z ) = 0 1.1. PAGES 2-3 5 (viii) 1+ i √ 2 n for 2 ≤ n ≤ 8 n = 2 : Re( z ) = 0 and Im( z ) = 1 n = 3 : Re( z ) =- √ 2 / √ 2 and Im( z ) = √ 2 / √ 2 n = 4 : Re( z ) =- 1 and Im( z ) = 0 n = 5 : Re( z ) =- √ 2 / √ 2 and Im( z ) =- √ 2 / √ 2 n = 6 : Re( z ) = 0 and Im( z ) =- 1 n = 7 : Re( z ) = √ 2 / √ 2 and Im( z ) =- √ 2 / √ 2 n = 8 : Re( z ) = 1 and Im( z ) = 0 2. Find the absolute value and the conjugate of each of the following: (i) z 1 =- 2 + i | z 1 | = √ 5 and ¯ z 1 =- 2- i (ii) z 2 =- 3 | z 2 | = 3 and ¯ z 1 =- 3 (iii) z 3 = (2 + i )(4 + 3 i ) (2 + i )(4 + 3 i ) = 5 + 10 i | z 3 | = √ 125 = 5 √ 5 and ¯ z 1 = 5- 10 i (iv) z 4 = 3- i √ 2+3 i 3- i √ 2 + 3 i = 3 11 ( √ 2- 1)- i 11 ( √ 2 + 9) 6 CHAPTER 1. THE COMPLEX NUMBER SYSTEM | z 4 | = √ 110 11 and ¯ z 1 = 3 11 ( √ 2- 1) + i 11 ( √ 2 + 9) (v) z 5 = i i +3 i i + 3 = 1 + 3 i 10 | z 5 | = √ 10 10 and ¯ z 1 = 1- 3 i 10 (vi) z 6 = (1 + i ) 6 (1 + i ) 6 =- 8 i | z 6 | = 8 and ¯ z 1 = 8 (vii) z 7 = i 17 | z 7 | = 1 and ¯ z 1 =- i 17 =- i 3. Show that z is a real number iff z = ¯ z Proof: ⇒ ( z is a real number) Let z = x + iy for { x,t } ∈ R . By definition of the conjugate, we find ¯ z = x- iy . But since z is real, y = 0. Therefore z = x = ¯ z . ⇐ ( z = ¯ z ) By hypothesis we find z = x + iy = x- iy = ¯ z So 2 iy = 0 which gives y = 0. Therefore z = x is a real number. 4. If z and w are complex numbers, prove the following equations: Let z = x 1 + y 1 and w = x 2 + y 2 for { x 1 ,x 2 ,y 1 ,y 2 } ∈ R (i) | z + w | 2 = | z | 2 + 2Re[ z ¯ w ] + | w | 2 | z + w | 2 = | ( x 1 + iy 1 ) + ( x 2 + iy 2 ) | 2 = | ( x 1 + x 2 ) + i ( y 1 + y 2 ) | 2 = ( x 1 + x 2 ) 2 + ( y 1 + y 2 ) 2 = ( x 2 1 + y...
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