# HW1s - Key to Complex Analysis Homework 1 Chapter 1(in...

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Unformatted text preview: Key to Complex Analysis Homework 1 Chapter 1 (in Greene & Krantz’s book) 13. Give all possible polar forms of each number. solution. (a) z = √ 3 + i : r = | z | = √ 3 + 1 = 2 θ = arg( z ) = arctan( = z < z ) + 2 kπ = π 6 + 2 kπ, k ∈ Z Hence, all the possible polar form of z can be written as: z = re iθ = 2 e π 6 i +2 kπi , k ∈ Z . (d) z = 4- 8 i : r = | z | = √ 16 + 64 = 4 √ 5 , θ = arg( z ) = arctan( = z < z ) + 2 kπ =- arctan(2) + 2 kπ, k ∈ Z Hence, all the possible polar form of z can be written as: z = re iθ = 4 √ 5 e- arctan(2) i +2 kπi , k ∈ Z . 2 14. Convert the polar forms into rectangular forms. solution. (b) z = 4 e iπ/ 4 Since r = 4 ,θ = π/ 4, we have that x = r cos( θ ) = 4 * √ 2 2 = 2 √ 2 y = r sin( θ ) = 4 * √ 2 2 = 2 √ 2 Hence, z = 2 √ 2 + 2 √ 2 i. (g) Since r = 7 ,θ =- π/ 6, we have that x = r cos( θ ) = 7 * √ 3 2 = 7 √ 3 2 y = r sin( θ ) =- 7 * 1 2 =- 7 2 Hence, z = 7 √ 3 2- 7 2 i. 1 2 28. Compute each of the following derivatives: (a) ∂ ∂z ( 4¯ z 2- z 3 ) (b) ∂ ∂ ¯ z ( ¯ z 2 + z 2 ¯ z 3 ) (c) ∂ 3 ∂x 2 ∂y ( 3 z 2 ¯ z 4- 2 z 3 ¯ z + z 4- ¯ z 5 ) (d) ∂ 5 ∂ ¯ z 3 ∂z 2 ( ¯ z 2- z ¯ z + 4 z- 6 z 2 ) Solution. We use the fact that ∂ ∂z z = 1 , ∂ ∂z ¯ z = 0 , ∂ ∂ ¯ z z = 0 , ∂ ∂ ¯ z ¯ z = 1 , as well as the usual sum and product rule for differentiation....
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HW1s - Key to Complex Analysis Homework 1 Chapter 1(in...

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