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Sol_HW1 - Math Methods Fall 2010 Solutions for Assignment 1...

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Math Methods Solutions for Assignment 1 Aug. 25, 2010 Fall 2010 Functions of Complex Variable Due Sep. 01, 2010 1 (6). Evaluate 1 + i = (1 + i ) 1 / 2 , 1 + i = 2 e iπ/ 4 , (1 + i ) 1 / 2 = 4 2 e iπ/ 8 . f ( z ) = z 1 / 2 is two-to-one mapping The second solution is 4 2 e iπ/ 8+ = 4 2 e 9 πi/ 8 . log(1 + i ) = log( 2 e iπ/ 4 ) = 1 2 log | 2 | + 4 + 2 πik . tan(1 + i ) = e i (1+ i ) - e - i (1+ i ) i ( e i (1+ i ) + e - i (1+ i ) ) = i e 1 - i - e i - 1 e i - 1 + e i - 1 = i e 2 - e 2 i e 2 + e 2 i . 2 (3). Explain the ‘paradox’: e = e 1+2 πi , and hence e = ( e 1+2 πi ) (1+2 πi ) = e 1 - 4 π 2 . f ( z ) = e z is multi-valued function. In general, ( e a ) b 6 = e ab , .e.g. ( e z ) 1 /n 6 = e z/n 3 (3). Show that f ( z ) = ¯ z is not differentiable. For the f ( x + iy ) = x - iy , the Cauchy-Riemann equations are not satisfied ∂u ( x, y ) ∂x = 1 6 = ∂v ( x, y ) ∂y = - 1 . 4 (3). Classify all singularities of f ( z ) = z sin z . z = 0 is a removable singularity, since lim z 0 z sin z = lim z 0 1 cos z = 1 exists. z = πn are poles of order 1, since lim z ( z - ) f ( z ) = lim z ( z - ) z sin z = lim z 2 z - cos z = ( - 1) n 6 = 0.

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