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Unformatted text preview: MATH 6322, COMPLEX ANALYSIS Complex numbers : Motivating problem: you can write down equations which dont 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 its 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 i (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 i 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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 Summer '08
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 Equations, Complex Numbers

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