math185-hw1sol - MATH 185 COMPLEX ANALYSIS FALL 2007\/08 PROBLEM SET 1 SOLUTIONS Throughout the problem set i =-1 and whenever we write i it is implicit

# math185-hw1sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 1 SOLUTIONS Throughout the problem set, i = - 1; and whenever we write α + βi , it is implicit that α, β R . 1. Determine the values of the following (without the aid of any electronic devices). (a) (1 + i ) 20 - (1 - i ) 20 . Solution. (1 + i ) 20 - (1 - i ) 20 = [(1 + i ) 2 ] 10 - [(1 - i ) 2 ] 10 = (2 i ) 10 - ( - 2 i ) 10 = 0 . (b) cos 1 4 π + i cos 3 4 π + · · · + i n cos( 2 n +1 4 ) π + · · · + i 40 cos 81 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 0 = a 2 = · · · = a 40 , a 1 = a 3 = · · · = a 39 , and a 0 + a 1 + · · · + a 40 = 21 a 0 + 20 a 1 = 2 2 (21 - 20 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. Use the exponential form of cos θ and sin θ to show the following. (a) Show that 1 + n cos θ + · · · + n ! r !( n - r )! cos + · · · + cos = (2 cos 1 2 θ ) n cos 1 2 nθ. Prove that, as n → ∞ , the series converges to 0 if 2 3 π < θ < 4 3 π . Date : September 19, 2007 (Version 1.0). 1
Solution. Using the form cos = 1 2 ( e irθ + e - irθ ), series becomes n X r =1 n r cos = 1 2 n X r =1 n r e irθ + 1 2 n X r =1 n r e - irθ = 1 2 (1 + e ) n + 1 2 (1 + e - ) n = 1 2 e i 1 2 ( e - i 1 2 θ + e i 1 2 θ ) n + 1 2 e - i 1

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