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# ComplexI - Calculus IVA Complex variables Lecture I...

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Unformatted text preview: Calculus IVA Complex variables. Lecture I October 28, 2003 Contents 1 Introduction 1 2 Relation with the partial derivatives 4 3 Rules of differentiation 7 4 The logarithm 9 5 Harmonic functions 10 6 Complex power series 12 7 Exercises 14 1 Introduction So far we have considered functions of a real variable with complex values. An example is the function e it = cos( t ) + i sin( t ) or more generally, if a is any complex number, the function g ( t ) = e at . 1 If a = p + iq then g ( t ) = e pt cos( qt ) + ie pt sin( qt ) . Similarly, we may consider functions of a complex variable with complex values. For instance, consider the function f ( z ) = z 2 . If we write z = x + iy then f ( z ) = ( x + iy ) 2 = x 2 + ( iy ) 2 + 2 ixy = x 2- y 2 + 2 ixy . Similarly, consider the function g ( z ) = z 2 defined by g ( z ) = z 2 = ( x- iy ) 2 = x 2- y 2- 2 ixy . An important example is the exponential function e z = e x e iy = e x cos( y ) + ie x sin( y ) . In general if F is a complex function of a complex variable we may write F ( z ) = P ( x, y ) + iQ ( x, y ) where P, Q are real valued functions of two variables. The function P is the real part of F and the function Q the imaginary part. We say that a function F ( z ) is holomorphic is for every z (in the domain of definition of F ) the following limit exists lim z 1 → z F ( z 1 )- F ( z ) z 1- z . This can also be written as lim h → F ( z + h )- F ( z ) h or lim Δ z → F ( z + Δ z )- F ( z ) Δ z . Assuming that F is holomorphic we define the holomorphic derivative of F to be the limit: F ( z ) = lim z 1 → z F ( z 1 )- F ( z ) z 1- z . 2 We also write dF dz = F ( z ) . To make the definition precise, we have to assume that for every z in the domain of definition D of F , there is a disc of center z of small enough radius contained in D : then z + Δ is still contained in D , provided | Δ z | is small enough. A domain with this property is said to be open. Example 1 : Check that z 2 is holomorphic at any point. Solution ( z + Δ z ) 2- z 2 Δ z = z 2 + 2 z Δ z + Δ z 2- z 2 Δ z = 2 zh + h 2 h = 2 z + h. As h → 0 this has a limit, namely 2 z . Thus this function is holomorphic and ( z 2 ) = 2 z . Example 2 : Check that F ( z ) = 1 1- z is holomorphic in the domain z 6 = 1. Solution Indeed F ( z + Δ z )- F ( z ) Δ z = 1 1- z- Δ z- 1 z Δ z = 1- z- (1- z- Δ z ) (1- z- Δ z )(1- z )Δ z = 1 (1- z- Δ z )(1- z ) Thus lim Δ z → F ( z + Δ z )- F ( z ) Δ z = lim Δ z → 1 (1- z- Δ z )(1- z ) = 1 1- z 2 . Thus F is indeed holomorphic and F ( z ) = 1 (1- z ) 2 . 3 Example 3 : Check that F ( z ) = e z is holomorphic. Solution We first check that the derivative exists at 0, that is, we show lim z → e z- e z = lim z → e z- 1 z exists. Since e z = X n ≥ z n n !...
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ComplexI - Calculus IVA Complex variables Lecture I...

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