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math185f08-hw4sol

# math185f08-hw4sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 4 SOLUTIONS For z C , recall that the argument of z , denoted arg( z ), is any θ R such that z = | z | e . Note that arg( z ) is only defined modulo 2 π . 1. This problem shows that the conditions in Theorem 2.8 in the lectures cannot be omitted. (a) Let f : C C be defined by f ( z ) = ( exp( - z - 4 ) z 6 = 0 , 0 z = 0 . Show that f satisfies the Cauchy-Riemann equation at all z C but f is not analytic on C . Why doesn’t this contradict the second part of Theorem 2.8 ? Solution. For z = x + iy C × , it follows from chain rule that f x ( z ) = 4 z - 5 exp( - z - 4 ) , f y ( z ) = 4 iz - 5 exp( - z - 4 ) . Hence the Cauchy-Riemann equation f x ( z ) = if y ( z ) is satisfied for all z C × . To compute f x (0) and f y (0), we need to use the definition of partial derivatives. f x (0) = lim ξ 0 R f (0 + ξ ) - f (0) ξ = lim ξ 0 R exp( - ξ - 4 ) - 0 ξ = lim s →∞ ,s R s exp( s 4 ) = 0 . This follows from setting s = 1 and using the exponential inequality e x 1 + x for a real variable x to get 0 s exp( s 4 ) s 1 + s 4 = 1 s - 1 + s 3 0 as s → ∞ . The same argument together with the observation that i 4 = 1 gives f y (0) = lim η 0 R f (0 + ) - f (0) η = lim η 0 R exp( - η - 4 ) - 0 η = lim s →∞ ,s R s exp( s 4 ) = 0 . Hence the Cauchy-Riemann equation is also satisfied at z = 0 since f x (0) = 0 = if y (0) . Date : October 24, 2008 (Version 1.0). 1

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However f is not analytic on C since it is not analytic at z = 0. In fact, it is not even continuous at z = 0. To see this, write z = r (cos θ + i sin θ ) and use De Moivre’s formula to get exp( - z - 4 ) = exp( - r - 4 cos 4 θ + ir 4 sin 4 θ ) = exp( - r - 4 cos 4 θ ) exp( ir 4 sin 4 θ ); if we consider the sequence z n := 1 n cos π 8 + i sin π 8 , n N , then z n 0 but f ( z n ) = exp( - z - 4 n ) = exp( in - 4 ) 9 0 since | exp( in - 4 ) | = 1 for all n N . So f is not continuous at 0.
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