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HW7sol - P ROBLEM SET 7 Partial solution Sec 2.4 p 2 Let...

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P ROBLEM SET 7 Partial solution Sec 2.4 p. 2 Let f ( z ) = ( e z - 1) 2 . Note that f ( z ) = 0 when e z = 1 . Writing z = x + iy gives e x e iy = 1 . Since the modulus of e x e iy has to be 1 we get that e x = 1 which implies x = 0 . Thus e iy = 1 and hence y = 2 πk where k is an integer. So there ane infinitely many zeroes z k = 2 πki , where k in an integer. Computing gives f ( z k ) = 0 , f ( z ) = 2( e z - 1) e z , f ( z k ) = 0 , f ( z ) = 4 e 2 z - 2 e z , and f ( z k ) = 2 . Since f ( z k ) = f ( z k ) = 0 and f ( z k ) = 0 , f has a zero of order 2 at z k for all k ’s. Another way to solve the problem is to show that e z - 1 has a zero of order 1 at z k and than use the fact that if a function g has zero of order m at z 0 then g 2 has zero of order 2 m at z 0 . This was a part of problem 17 on the problem set. Sec 2.4 p. 10 Let f ( z ) = e z and z 0 = πi . By Taylor’s expansion e z = k =0 f ( k ) ( z 0 ) k ! ( z - z 0 ) k Note that f ( k ) ( z ) = e z and hence f ( k ) ( πi ) = e πi = - 1 . Thus e z = - k =0 1 k ! ( z - πi ) k . Since e z is entire by Theorem 1 Sec 2.4 (a consequence of Cauchy’s Formula that gives
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