## Slide 2 Cauchy Theorem and Integral formula

Relating the kth derivative and the integral over a closed curve $\gamma$ of a function of the form $\int_{\gamma}\frac {f(z)}{{(z-w)}^{k+1}}$

D is a simply connected domain (no holes), F is analytic in a region that contains D, then,

$f^{(k)}(w) = \frac{k!}{2 \pi i}\int_{\gamma}\frac {f(z)}{{(z-w)}^{k+1}}dz$ for all w in D, k > 0.

## Slide 3 Cauchy’s estimate

Slide 4 Liouville’s Theorem

f is analytic in the complex plane (i.e. entire function)

and if f is bounded

then f must be constant.

converse : There are no bounded, varying analytic functions in C.

c.f. Sin x in reals where function value is bounded between -1 and +1 but is (definitely) not constant.

Proof: uses Cauchy’s estimate

## Slide 7 Consequence of fundamental theorem of algebra

A polynomial can be factored into n factors.

## Slide 8 Maximum Principle *Important *

Let f be analytic in domain D

and suppose there exists a point z_0 in D such that $f(z) \leq f(z_0)$ for all z in D

i.e. there is a maximum value at some point in D,

then f is constant in D.

Consequence if f is analytic on D then |f(z)| reaches its maximum on the boundary $\partial D$

## Slide 9 Maximum Principle Example *Important *

<< diagram >>

Domain is square where one corner is origin other corners are 1, i and 1+i, so sides of length 1. z = x + iy, so both x and y are real values between 0 and 1.

$f(z) = z^2 - 2z$ (which is analytic but not constant)

What is the maximum of |f(z)|?

Boundary $\gamma_1 , 0 \leq x \leq 1, y = 0$

so |f(z)| = |f(x)|= |x^2 – 2x| = |x(x-2)|

for the domain of x, maximum occurs when x = 1, so $|f(z)| \leq |f(1)| = 1 \textrm{ on } \gamma_1$

Boundary $\gamma_2 , 0 \leq y \leq 1, x = 1$

so |f(z)| = |f(1+iy)|= $|-1-y^2| = y^2 + 1$

for the domain of y zero to 1 and fixed value of x=1, maximum occurs when y = 1, so $|f(z)| \leq |f(1+i)| = 2 \textrm{ on } \gamma_2$

<< provide detailed working for further two boundaries >>

$\gamma_3 \; y = 1, 0 \leq x \leq 1$

By considering the modulus of the function applied to x + i (which is one way of describing this boundary) and using pythag to get expression for the modulus of f(x+i)

e.g. wolfram gives $max((x^2 -2x -1)^2 +(2x-2)^2)^{0.5}, \textrm { for x=0 to 1, is } \sqrt{5} \textrm{ at } x = 0$

$\gamma_4 \; x = 0, 0 \leq y \leq 1$