# Conformal mappings on the complex plane

## Homework Statement

a) Using the conformal mapping w=cosh(z), find a rectangle R in the z-plane which maps to the region in the w-plane with boundaries as follows:

- a plate of constant temperature on the line segment {w=u+iv : -1<u<1, v=0}
- an outer boundary of cooler constant temperature given by the ellipse u^2/cosh^2(1) + v^2/sinh^2(1) = 1

b) The elliptical surface is at temperature 0 and the line segment is at tempertaure 1. Choose a complex function g, defined on R, such that the real part of g(inv_cosh(w)_ defines a temperature with appropriate boundary conditions in the w-plane.

## Homework Equations

z=x+iy
cosh(z) = (e^z+e^-z)/2 = ... = cosh(x)cos(y) + i*sinh(x)sin(y)
general equations for ellipse and hyperbola

## The Attempt at a Solution

Firstly I drew the plate and outer boundary.
To map into a rectangle in the z-plane I first recognise that hyperbolae transform to horizontal lines through w=cosh(z) and similarly ellipses transform to vertical lines.
With this in mind I can recognise that the 'plate' is simply a hyperbolae with v=0.
So for the plate
v=sinh(x)sin(y)=0
u=cosh(x)cos(y)=-1...1

sin(y) is not = 0 because then cos(y)=1 and hence cosh(x)=-1...1
Therefore sinh(x)=0
x=0

u=cosh(0)cos(y)=-1...1
y=inv_cos(-1...1)
=0...Pi

So we have mapped the plate to z=y, 0<y<Pi
But my problem here is this is sinusoidal so really n*Pi, i.e. an infinite line?
Also to make my rectangle R of boundary conditions won't I need two horizontal lines? Are these y=0 and y=Pi?

Anyway my attempt on transforming the ellipse to two vertical lines is similar:
cosh^2(1)=2.381
sinh^2(1)=1.381

u^2/2.381 + v^2/1.381 = 1
cosh^2(x)cos^2(y)/2.381 + sinh^2(x)sin^2(y)/1.381 = 1

Presumably y=0 but only because I know a priori that it should map to vertical lines

Anyway... cosh^2(x) = 2.381
x=inv_cosh(+-1.543)
=~2.5
only 1 value :(

I think my methods are wrong but don't know what else to do. (I tried starting the map from the other direction, i.e. from lines to ellipses/hyperbolae but got confused with the multiple values).
Thanks.

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Dick
Homework Helper
You have that cosh(z) defines the conformal mapping u=cosh(x)cos(y), v=sinh(x)sin(y). Compare this to the parametric form for an ellipse centered at the origin with semimajor axes r_u and r_v. u=r_u*cos(t), v=r_v*sin(t) for t=0 to 2pi. There aren't any hyperbolae in the problem at all. All you have to do is figure out the correct x and y intervals defining the rectangle.

So I set the two forms equal and eliminate y to obtain
1 = (r_u*cos(t)/cosh(x))^2 + (r_v*sin(t)/sinh(x))^2

where I know r_u=cosh(1) and r_v=sinh(1) looking at my specific ellipse

and somehow solve this to get my two values for x?

Dick
Homework Helper
I would say r_u=cosh(x) and r_v=sinh(x) and y=t. Can you figure out the x and y ranges of the rectangle now? You want r_u to go from 1 to cosh(1) and r_v to go from 0 to sinh(1). The suggestion w=cosh(z) basically gave you everything. You just have to extract the boundaries of the rectangle from it.

Why from 1 to cosh(1)?
Is it because we want the gap between the plate and boundary to form our rectangle boundaries?

Dick
Homework Helper
Because the ellipse starts with a u semimajor axis of 1 and ends with a semimajor axis of cosh(1). Draw a picture! That's cosh(0) to cosh(1).

We have
u=cosh(x)cos(y)
v=sinh(x)sin(y)

and we want u to vary from 1 to cosh(1) and v to vary from o to sinh(1)

so I want to solve the four equations
cosh(x)cos(y)=1
sinh(x)sin(y)=cosh(1)
cosh(x)cos(y)=0
sinh(x)sin(y)=sinh(1)
to get my four boundary lines to the rectangle?

I think I am making this too complicated!

Dick