## Coursera: Analysis of a Complex Kind. 4.2 Conformal Mappings

### Jul 24, 2016

Basic Idea: A conformal mapping, f(z), preserves the angle between two curves that intersect in the z-plane.

First a large number of definitions are required.

## Slide 3 Path

A path from (complex point) A to (complex point) B is a continuous function that maps the (real) interval [a,b] to $\gamma{(a)} \textrm{ to } \gamma{(b)}$. A bit like the graph that is created when draw (x,y) from real-valued y=f(x).

e.g. $\gamma{(t)}$ = (2 + i) + $e^{it}, 0 \leqslant t \leqslant \pi$, is a unit half-circle centre 2 + i.

$\gamma{(a)} \textrm{ and } \gamma{(b)}$ are complex numbers, and as t moves from (real) a to b, if gamma is continuous, the resulting path could be a straight line, a curve or some looping path.

<< image here >>

## Slide 5 Tangent vector on a path

<< sketch here >>

if $\gamma = x + iy:[a,b] \to C$ is a smooth curve, and $t_0$ is some value on the interval [a,b], then the derivative

$\gamma ' (t_0) = x'(t_0) + iy'(t_0)$

is a tangent vector to $\gamma \textrm{ at } z_0 = \gamma(t_0)$

## Slide 6 angle between curves

<< sketch here >>

If two curves intersect at $z_0$ then angle between the curves is the angle between the tangent vectors. Note the “sense” of the curves as the parameter increases, in what direction are you travelling along the path?

<< algebra here >>

## Slide 7  Example angle between curves at a point

We have two curves $\gamma_1 \, \gamma_2$, parts of circles, that intersect at one point $z_0$

Equations and simple sketches of curves

Now for derivatives

## Slide 8 Definition for conformal mapping

A function is conformal if it preserves angles between curves.

If curves $\gamma_1 \textrm{ and } \gamma_2$ have angle alpha between them over single point or a whole domain D (see slide 11 for the significance of this), then the function g (if continuous) is conformable if it preserves angles between $g({\gamma_1}) \textrm{ and } g({\gamma_2})$

## Slide 9 conditions and justification for conformal mapping

If f:U -> C is analytic  and if $z_0 \; \in$ U such that $f \, ' \, (z_0) \neq 0$ then  f is conformal at $z_0$

<< to do justification >>

## Slide 11 f(z) = $e^z$

The exponential function has the interesting property that for each point in C it is conformal, but it is not conformal for the whole of C as the domain.

### Justification

f is analytic in C and f ‘ (z) $\neq$ 0 for any point in C. Therefore $e^z$ is conformal for each point in C.

However f(z) = $e^z$ is not one-one (*) in C it is not a conformal mapping onto C\{0}.

($e^z$ has period $2i\pi$ )

Now, if we restrict the domain the exponential applies to D={z| 0 < Im z < 2π} (a horizontal strip of width 2π) then f is now one-one and it maps conformally onto f(D) = C\[0,\infty)

(because the domain, D, does not include the real axis, the exponential cannot reach the real axis, but it can reach everywhere else in C)

Visualisation of exponential. For each horizontal line in domain, with constant (k) imaginary value, and variable real value, x,

exp(x +ik) = e^x(cos k + i sin k)

E^x varies in length but always positive

(cos k + i sin k) is a point on unit circle with argument k (fixed)

so the function maps to lines radiating out from origin with argument k that lies between 0 and 2π