math185f08-hw1sol

# math185f08-hw1sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 1 SOLUTIONS Throughout the problem set, i = √- 1; and whenever we write a + bi , it is implicit that a,b ∈ R . 1. Determine the values of the following (without the aid of any electronic devices). (a) (1 + i ) 200- (1- i ) 200 . Solution. (1 + i ) 200- (1- i ) 200 = [(1 + i ) 2 ] 100- [(1- i ) 2 ] 100 = (2 i ) 100- (- 2 i ) 100 = 0. (b) cos 1 4 π + i cos 3 4 π + ··· + i n cos( 2 n +1 4 ) π + ··· + i 400 cos 801 4 π Solution. Write a n = i n cos( 2 n +1 4 ) π . Note that a n +2 =- i n cos ( 2 n +1 4 ) π + π = i n cos( 2 n +1 4 ) π = a n . So a = a 2 = ··· = a 400 , a 1 = a 3 = ··· = a 399 , and a + a 1 + ··· + a 400 = 201 a + 200 a 1 = √ 2 2 (201- 200 i ) . (c) 1 + 2 i + 3 i 2 + ··· + ( m + 1) i m where m is divisible by 4. Solution. Let S be the sum. Then S = 1 + 2 i + 3 i 2 + ··· + ( m + 1) i m , iS = i + 2 i 2 + ··· + mi m + ( m + 1) i m +1 . Subtracting the second equation from the first yields (1- i ) S = 1 + i + i 2 + ··· + i m- ( m + 1) i m +1 = 1- i m +1 1- i- ( m + 1) i m +1 = 1- ( m + 1) i since i m = 1 if m is divisible by 4. Hence S = 1- ( m + 1) i 1- i × 1 + i 1 + i = 1 2 ( m + 2- mi ) . 2. Let z ∈ C be such that Im( z ) 6 = 0 and Im 1 + z + z 2 1- z + z 2 = 0 . Prove that | z | = 1. Solution. Note that 0 = Im 1 + z + z 2 1- z + z 2 = Im 1 + 2 z 1- z + z 2 = 2Im z 1- z + z 2 . So z 1- z + z 2 ∈ R . Since Im( z ) 6 = 0, this implies z 6 = 0, and so 1- z + z 2 z ∈ R , Date : September 26, 2008 (Version 1.0). 1 ie. 1 z- 1 + z ∈ R , ie. z + 1 z ∈ R , ie. z + 1 z = ¯ z + 1 ¯ z . Clearing denominators from the last equation, we get z 2 ¯ z + ¯ z = z 2 z + z. Since z ¯ z = | z | 2 , we get ( z- z )( | z | 2- 1) = 0 . Hence either z = z or | z | 2 = 1. Since Im( z ) 6 = 0, z 6 = ¯ z , and so we must have | z | = 1 ....
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math185f08-hw1sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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