review1 - MATH 6322, COMPLEX ANALYSIS, Reivew 1 In the...

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Unformatted text preview: MATH 6322, COMPLEX ANALYSIS, Reivew 1 In the following, U ⊂ C is always assumed to be (connected and) open. Holomorphic Functions : • Theorem( Alternative definitions of holomrphic functions on U ): Let f : U → C be C 1 ( U ). Then the followings are the equivalent definitions that f is holomr- phic on U : 1. ∂f ∂ ¯ z = 0 on U ; 2, f satisfies Cauchy-Riemann equations on U ∂u ∂x = ∂v ∂y , ∂u ∂y =- ∂v ∂x when we write f = u + iv on U ; 3. f ( z ) exists for every z ∈ U , i.e. f is complex differentiable on every points z ∈ U ; 4. f can expanded locally as a power series ∑ ∞ n =1 a n ( z- z ) n at every point z ∈ U ; 5. For every piecewise C 1 curve γ : [0 , 1] → U with γ (0) = γ (1) (so γ is closed), R γ f ( z ) dz = 0. Idea of pf : (1) ⇐⇒ (2) is trivial by definition that ∂ ∂ ¯ z = 1 2 ( ∂ ∂x + i ∂ ∂y ); (1) ⇐⇒ (3) uses f ( z )- f ( z ) = ∂f ∂z ( z- z ) + ∂f ∂ ¯ z (¯ z- ¯ z ) + o (( z- z )); (1) = ⇒ (5) is actually the Cauchy’s theorem whose proof use the existence of the primitive (anti-derivative) of f , and (5) = ⇒ (1) is the so-called Morera’s theorem, its proof is to use the condition that R γ f ( z ) dz = 0 to define a F with F = f . (4) = ⇒ (1) is trivial, while (1) = ⇒ (4) uses Cauchy’s formula f ( z ) = 1 2 πi Z ∂D ( P,r ) f ( ζ ) ζ- z dζ, and the fact that, for | ζ- P | = r and | z- P | < r , 1 ζ- z = 1 ζ- P 1 1- z- P ζ- P = ∞ X n =0 ( z- P ) n ( ζ- P ) n +1 . 1 • Goursat’s theorem . If f has complex derivative at each point in U (where U is an open set in C ), then f is holomorphic. Idea of pf : Use Morera’s theorem. • Practice Problems : (1) If f and ¯ f are both holomorphic on a connected open set U ⊂ C , then prove that f is identically constant. Brief Solution : Show that ∂ ∂z f ≡ 0 and ∂ ∂ ¯ z f ≡ 0 on U . (2) Let f be holomorphic on U . Assume that | f ( z ) | is constant, then f is also a constant. Brief Solution : At least two methods: Method 1 : Prove, ∂ ∂z f ≡ 0 and ∂ ∂ ¯ z f ≡ (automatic since f is holomorphic) on U . Method 2 : Use “open mapping theorem”. (3) Let f be holomorphic on U . Assume that f be real valued, i.e. Im ( f )( z ) = 0 for all z ∈ U . Prove that f is constant (Hint: Use CR equation). Brief Solution : Use Cauchy-Riemann equation. (4) Defin4 f ( z ) = e z 3- 1 z for z 6 = 0 and f (0) = 0. Prove that f is an entire function, i.e. f is holomorphic on C . Brief Solution : Show that f (0) exists. Also observe, trivially, f ( z ) exists for all z other than 0 since f ( z ) = e z 3- 1 z for z 6 = 0. By Goursat’s theorem, f is entire. (5) Let f be holomorphic on C and define f * ( z ) = f (¯ z ). Prove that f * is holomorphic on C ....
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This note was uploaded on 02/21/2012 for the course MATH 4377 taught by Professor Staff during the Summer '08 term at University of Houston.

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review1 - MATH 6322, COMPLEX ANALYSIS, Reivew 1 In the...

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