# ans4 - Math 185 Sample Answers to Problem Set#4 Page 89 1a...

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Unformatted text preview: Math 185. Sample Answers to Problem Set #4 Page 89 1a. exp(2 ± 3 πi ) = e 2 (cos( ± 3 π )+ i sin( ± 3 π )) = e 2 (- 1+0 i ) =- e 2 (since cos( ± 3 π ) = cos π =- 1 and similarly for sin( ± 3 π ) ). 6. By (7), we have | exp( z 2 ) | = exp(Re( z 2 )) = exp( x 2- y 2 ) ≤ exp( x 2 + y 2 ) = exp( | z | 2 ) . Page 94 10. Way #1. Let u ( x, y ) = ln( x 2 + y 2 ) . Then u x = 2 x x 2 + y 2 ; u xx = 2( x 2 + y 2 )- (2 x ) 2 ( x 2 + y 2 ) 2 = 2( y 2- x 2 ) ( x 2 + y 2 ) 2 ; u y = 2 y x 2 + y 2 ; u yy = 2( x 2 + y 2 )- (2 y ) 2 ( x 2 + y 2 ) 2 = 2( x 2- y 2 ) ( x 2 + y 2 ) 2 =- u xx . Therefore u xx + u yy = 0 , and also we see from the above that all second-order partial derivatives of u are continuous on the domain of u , so u is harmonic. Way #2. Working as on page 78, we find that a candidate for a harmonic conjugate of u is v ( x, y ) = 2 tan- 1 ( y/x ) . Then a candidate for a holomorphic function with real part u ( x, y ) is f ( x + iy ) = ln( x 2 + y 2 ) + 2 i tan- 1 ( y/x ) = ln r 2 + 2 iθ = 2(ln r + iθ ) = 2 log z . Now we start the actual proof. Letting f ( z ) = log z 2 , we have u ( x, y ) = Re log z 2 , so it is harmonic....
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ans4 - Math 185 Sample Answers to Problem Set#4 Page 89 1a...

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