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

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Unformatted text preview: 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 := { z C | Re( z ) < } . Let z and let z be the line segment from 0 to z . Then by Proposition 3.9 , Z z exp( w ) dw = exp( z )- exp(0) = e z- 1 . By Proposition 3.5 , Z z exp( w ) dw sup w z | exp( w ) | Z z | dw | = | z | sup w z e Re( w ) | z | where the last step follows since z 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 . Now observe that Z 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 z exp( w )- f n- 1 ( w ) dw sup w z | exp( w )- f n- 1 ( w ) | Z z | dw | | z | sup w 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 (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)?...
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This note was uploaded on 06/15/2009 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.

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

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