Coursera Analysis Complex Kind Lesson 5.4 Cauchy’s Theorem

Aug 3, 2016

The BIG one !!

Slide 2 Cauchy’s theorem statement and proof strategy

D is a simply connected domain in C.

(function ) f is analytic in D.

\gamma is a piecewise , smooth, closed curve in D i.e. \gamma : [a,b] \to D \;  , \bf{\gamma(b) \; = \; \gamma(a)}

the \int_{\gamma}f(z) \; dz = 0

For example f(z) = e^{z^3} is analytic (everywhere) in C and C is simply connected.

Now it is difficult (impossible?) to find a primitive of e^{z^3}

But for any closed curve, \gamma ,  in C, \int_{\gamma}e^{z^3} \; dz = 0

Proof approach

Since D has no holes we can deform \gamma to a point in D.

We have to show the integral does not change as this deformation occurs.

Slide 3 Corollary of Cauchy theorem

Slide 4 example

<< diagram >>

R is rectangle centred z_0.

Claim \int_{\partial R}\;\frac{1}{z-z_0} \;dz = 2i\pi, where \partial R is the boundary of the rectangle. Notice z = z_0 is excluded from the domain.

Proof: Look at the circle that surrounds the rectangle and touches it at some points (is this necessary?), | z – z_0| = r

<< diagram >>

Now \int_{| z -  z_0| = r}\;\frac{1}{z-z_0} \;dz = 2i\pi

** learn following **

First parameterise the path, \gamma \; , | z -  z_0| = r

Well this is circle centre z_0, \textrm{radius r = }z_0 + re^{it} , \; 0\leq t < 2 \pi

So the integrand becomes: \frac{1}{z-z_0} \; = \; \frac{1}{z_0 + re^{it} - z_0} = \; \frac{1}{ re^{it}}

Further \gamma \; ' \; = \;ire^{it}

Merging: \int_0^{2 \pi} \;  \frac{1}{ re^{it}}.ire^{it} \; dt \; = \; i\int_0^{2 \pi} \;   dt \; = \; 2i \pi

and by the Cauchy Theorem this is also the value of the integral along the boundary of the rectangle within the circle.

Slide 5 Example

** learn following example **

\int_{|z|=1}\frac{1}{z^2 + 2z}dz

First, partial fractions = \frac{1}{2}(\int_{|z|=1}\frac{1}{z}dz - \int_{|z|=1}\frac{1}{z + 2}dz) (*)

<< diagram >>

|z| = 1 is circle centre 0, radius = 1.

1/z has excluded point z = 0 in |z| = 1, and we calculated previously = 2i \pi

1\z + 2, excluded point z = -2, is NOT inside path |z| = 1 therefore integral around closed loop = 0

So (*) becomes \frac{1}{2}(2i \pi - 0)\; = i \pi

Slide 6 Cauchy Integral Formula (important technique for calculating integrals)

<< Diagram >>

D simply connected domain

boundary of D is piecewise smooth (closed) curve \gamma .

f is analytic in U which contain the closure of D (i.e. D and \gamma )

>> then f(w) = \frac {1}{2i \pi}\int_{\gamma}\frac{f(z)}{z-w}\;dz <<

Slide 7 Proof of Cauchy Integral Formula

Slide 8 Example

\int_{|z|=2}\frac{z^2}{z-1}dz

Note integrand is of form \frac{f(z)}{z-w}

Next question, does w=1 lie inside the path |z| = 2? YES.

So from integral formula : \frac {1}{2i \pi}\int_{|z|=2}\frac{z^2}{z-1}\;dz\;= f(w) \; = \; f(1) \;=\; 1

Rearranging

\int_{|z|=2}\frac{z^2}{z-1}\;dz\;= {2i \pi}

Slide 9 Example similar to previous

\int_{|z|=1}\frac{z^2}{z-2}dz

Note, integrand is of “correct” form but w=2 is not inside the curve |z| = 1 and also the rational function is analytic everywhere inside disk B1.5(0) therefore integral around the closed curve |z| = 1 is zero, by Cauchy theorem.

Slide 10 example using Log function

<< continue >>

 

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