math185-hw5sol

math185-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2007/08 PROBLEM SET 5 SOLUTIONS Notations: D (0 , 1) = { z ∈ C | | z | < 1 } ; ∂D (0 , 1) = { z ∈ C | | z | = 1 } ; N = { 1 , 2 , 3 ,... } ; f ◦ g denotes the composition of f and g and is defined by f ◦ g ( z ) = f ( g ( z )). 1. Let f : C → C be an entire function. Let a ∈ R be an arbitrary constant. (a) Show that if Re f ( z ) ≤ a for all z ∈ C , then f is constant. (b) Show that if Re f ( z ) ≥ a for all z ∈ C , then f is constant. (c) Show that if [Re f ( z )] 2 ≤ [Im f ( z )] 2 for all z ∈ C , then f is constant. (d) Show that if [Re f ( z )] 2 ≥ [Im f ( z )] 2 for all z ∈ C , then f is constant. (e) Suppose h is another entire functions and suppose there exists an a ∈ R , a > 0, such that Re f ( z ) ≤ a Re h ( z ) for all z ∈ C . Show that there exist α,β ∈ C such that f ( z ) = αh ( z ) + β for all z ∈ C . [Hint: if f and g are both entire, then so are f ◦ g and g ◦ f ; find an appropriate g so that you may apply Liouville’s theorem.] 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 Re f ( z ) e i Im f ( z ) | = e Re f ( z ) ≤ e a . • For (b), we choose g ( z ) = e- z and note that | e- f ( z ) | = | e- Re f ( z ) e- i Im f ( z ) | = e- Re f ( z ) ≤ e- a ....
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This note was uploaded on 08/01/2008 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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math185-hw5sol - MATH 185 COMPLEX ANALYSIS FALL 2007/08...

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