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 4 example

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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

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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 = $\frac{1}{2}(\int_{|z|=1}\frac{1}{z}dz - \int_{|z|=1}\frac{1}{z + 2}dz)$ (*)

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|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 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.

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