Exam Solusions and Feedback 2014

# Exam Solusions and Feedback 2014 - MATH 20142 Complex...

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MATH 20142 Complex Analysis: Test Manchester, March 14, 2014 Answer ALL four questions Problem 1 Find all complex solutions of the following quadratic equation: z ( z + 2) = - 1 + 18 i Solution. We have ( z + 1) 2 = 18 i. Denote z + 1 = a + ib ; then a 2 + 2 iab - b 2 = 18 i , whence we have a system a 2 - b 2 = 0 , ab = 9 . Therefore, b = a (since b = - a yields a 2 = - 9, which is impossible), whence a = b = ± 3. Answer: z = 2 + 3 i or z = - 4 - 3 i . [5 marks] Problem 2 Let f ( z ) = | z | 2 + sin z. Write f in the form f ( x + iy ) = u ( x, y ) + iv ( x, y ). Solution. We have f ( x + iy ) = x 2 + y 2 + sin x cosh y + i cos x sinh y , whence u ( x, y ) = x 2 + y 2 + sin x cosh y, v ( x, y ) = cos x sinh y . [2 marks] Use the Cauchy-Riemann equations to determine all z C at which f is differentiable. Solution. We have ∂u/∂x = 2 x + cos x cosh y, ∂u/∂y = 2 y + sin x sinh y, ∂v/∂x = - sin x sinh y, ∂v/∂y = cos x cosh y, The Cauchy-Riemann equations: ∂u/∂x = ∂v/∂y , whence x = 0; ∂u/∂y = - ∂v/∂x , whence y = 0. Therefore, z = 0 is the only point at which f can be differentiable. Since all the partial derivatives are clearly continuous, f is indeed differentiable at 0. [3 marks]

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Problem 3 Let f ( z ) = n =1 2 n + 1 n z n . Determine the radius of convergence
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Unformatted text preview: Solution. From the lectures, R = lim n → + ∞ ± ± ± ± a n − 1 a n ± ± ± ± = lim n → + ∞ (2 n − 1 + 1) n ( n-1)(2 n + 1) = 1 2 . [2 marks] • Find a closed formula for f ′ ( z ) (without inﬁnite summation) for any z such that | z | < R . Solution. We have for | z | < 1 / 2, f ′ ( z ) = ∞ ∑ n =1 (2 n + 1) z n − 1 = 1 z ∞ ∑ n =1 2 n z n + ∞ ∑ n =1 z n − 1 = 1 z ( ∞ ∑ n =1 (2 z ) n ) + 1 1-z = 1 z ² 2 z 1-2 z ³ + 1 1-z = 3-4 z (1-z )(1-2 z ) . [3 marks] Problem 4 Find all complex solutions z of the equation sinh z =-i. Solution. We have sin( iz ) = i sinh z , whence sinh z =-i sin( iz ). Therefore, sin( iz ) = 1, whence iz = π/ 2 + 2 πk, k ∈ Z . Answer: z =-iπ/ 2 + 2 iπk, k ∈ Z . [5 marks] (Alternatively, one can write sinh z = ( e z-e − z ) / 2, get e z =-i and compute the logarithm.)...
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• Spring '14
• Sidorov
• Equations, Quadratic equation, Elementary algebra

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