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math185f08-hw5sol

math185f08-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2008/09...

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MATH 185: COMPLEX ANALYSIS FALL 2008/09 PROBLEM SET 5 SOLUTIONS 1. Consider the n th Taylor polynomial approximant to exp( z ), f n ( z ) := 1 + z + 1 2! z 2 + · · · + 1 n ! z n . Show that for all z C with Re( z ) < 0, | exp( z ) - f n ( z ) | ≤ | z | n +1 for all n N . Solution. Let Ω := { z C | Re( z ) < 0 } . Let z Ω and let Γ z be the line segment from 0 to z . Then by Proposition 3.9 , Z Γ z exp( w ) dw = exp( z ) - exp(0) = e z - 1 . By Proposition 3.5 , Z Γ z exp( w ) dw sup w Γ z | exp( w ) | × Z Γ z | dw | = | z | sup w Γ z e Re( w ) ≤ | z | where the last step follows since Γ z Ω and thus Re( w ) 1. Hence | e z - 1 | ≤ | z | . Suppose the statement is true for n - 1, ie. | exp( z ) - f n - 1 ( z ) | ≤ | z | n for all z Ω. Now observe that Z Γ z exp( w ) - f n - 1 ( w ) dw = exp( z ) - z + 1 2! z 2 + · · · + 1 n ! z n - exp(0) = exp( z ) - f n ( z ) . By Proposition 3.5 and the induction hypothesis, Z Γ z exp( w ) - f n - 1 ( w ) dw sup w Γ z | exp( w ) - f n - 1 ( w ) | × Z Γ z | dw | ≤ | z | sup w Γ z | w | n = | z | n +1 . Hence | exp( z ) - f n ( z ) | ≤ | z | n +1 as required. By mathematical induction, this holds for all n . 2. Let f : C C be an entire function. As usual, we write f ( z ) = u ( x, y ) + iv ( x, y ) for z = x + iy . (a) Show that if u is positive valued, ie. u ( x, y ) > 0 for all x, y R , then f is constant. Date : October 24, 2008 (Version 1.0). 1
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(b) Show that if | u ( x, y ) | < | v ( x, y ) | for all x, y R , then f is constant. (c) Can we still draw the same conclusions if ‘ < ’ is replaced by ‘ > ’ in (a) and (b)? Solution. Note that e x is a monotone increasing function on R . For (a), we choose g ( z ) = e z and note that | e f ( z ) | = | e u e iv | = e u e 0 = 1 .
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