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## 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.