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# runote - MATH 6322 COMPLEX ANALYSIS Complex numbers •...

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Unformatted text preview: MATH 6322, COMPLEX ANALYSIS Complex numbers : • Motivating problem: you can write down equations which don’t have solutions, like x 2 + 1 = 0. Introduce a (formal) solution i , where i 2 =- 1. Define the set C := { a + ib a,b ∈ R } . Can put a ring structure on this. ( a 1 + ib 1 ) + ( a 2 + ib 2 ) := ( a 1 + a 2 ) + i ( b 1 + b 2 ) , ( a 1 + ib 1 ) · ( a 2 + ib 2 ) := ( a 1 a 2- b 1 b 2 )) + i ( a 1 b 2 + a 2 b 1 ) . One verifies that it’s associative, distributive, and has the expected units. It is a field, and ( a + ib )- 1 = a a 2 + b 2- i b a 2 + b 2 . • Conjugate: ¯ z = a- ib if z = a + ib . Norm: | z | := √ a 2 + b 2 . • e a + ib := e a (cos b + i sin b ). We can write: z = | z | e iθ (this can also be seen that C = R 2 so that we can use the polar coordinate to represent ( a,b ). The express of z = | z | e iθ is called the polar form of z . Complex differential operators : • Write ∂ ∂z = 1 2 ∂ ∂x- i ∂ ∂y ! , ∂ ∂ ¯ z = 1 2 ∂ ∂x + i ∂ ∂y ! . • Let f = u + iv be a complex valued function on U which is C 1 . If u and v are differentiable on U , then, for fixed z ∈ U , f ( z )- f ( z ) = ∂f ∂z ( z- z ) + ∂f ∂ ¯ z (¯ z- ¯ z ) + o (( z- z )) . ( * ) Holomorphic Functions : • A complex valued function f on U is said to be holomorphic on U if f is C 1 ( U ) and ∂f ∂ ¯ z = 0 on U , or equivalently, it is C 1 ( U ) and it satisfies the Cauchy- Riemann equations on U ∂u ∂x = ∂v ∂y , ∂u ∂y =- ∂v ∂x when we write f = u + iv on U . 1 • f is said to be holomorphic at (or around) z if f is holomorphic in a neighbor- hood of z . • The concept of holomorphic function is also related to the complex differentia- bility of f : A function f defined around z is said to be complex differentiable at z if the complex derivative lim h → a f ( z + h )- f ( z ) h exists. Form (*) we get the following: Assume that f is C 1 ( U ) for U ⊂ C . Then f is holomorphic (i.e. ∂f ∂ ¯ z = 0 on U ) if and only if f ( a ) exists for every a ∈ U . Furthermore, f ( z ) = ∂f ∂z = ∂f ∂x =- i ∂f ∂y on U . • Warning : f ( z ) exists alone does not imply that f is holomorphic at z . For example, function f ( z ) = ¯ z 2 is differentiable at 0 and nowhere else, in particular, f is not holomorphic at 0. • Important : (i) Goursat actually proved that f is holomorphic if f has complex derivative at each point in U ( Without the assumption that f is C 1 ). The proof will be given later. • If f is holomorphic around P , then f is conformal at P (i.e. it preserves the angles and stretches in all directions (see Theorem 2.2.3 on Page 40). • Examples of holomorphic functions on C : e z := e x · e iy = e x (cos y + i sin y ) , The (complex) sine and cosine functions: sin z = 1 2 i ( e iz- e- iz ) , cos z = 1 2 ( e iz + e- iz ) ....
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runote - MATH 6322 COMPLEX ANALYSIS Complex numbers •...

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