Conformal Mapping: Find Function to Map Between Two Parallel Lines

[skrat]

Homework Statement

Find function that maps area between $|z|=2$ and $|z+1|=1$ on area between two parallel lines.

Homework Equations

The Attempt at a Solution

I don't know how to check if my solution works for this problem?

I used Möbious transformation:

$f(z)=\frac{az+b}{cz+d}$ where I decided that $f(-2i)=\infty$ and $f(2i)=0$ and $f(2)=2i$.

Using this I find out that $f(z)=\frac{(z-2i)(i-4)}{\frac{1-i}{2}z+i+1}$

Now I am almost 100% if the circles would both have center in point 0. However, they have different origins and I somehow don't know how to deal with this kind of problems. :(

 

[pasmith]

Homework Statement

Find function that maps area between $|z|=2$ and $|z+1|=1$ on area between two parallel lines.

Homework Equations

The Attempt at a Solution

I don't know how to check if my solution works for this problem?

I used Möbious transformation:

$f(z)=\frac{az+b}{cz+d}$ where I decided that $f(-2i)=\infty$ and $f(2i)=0$ and $f(2)=2i$.

The circles |z| = 2 and |z + 1| = 1 intersect only at z = -2, so you must have f(-2) = \infty.

Also, the first choice you need to make is the parallel lines bounding the image of the area in question.

Using this I find out that $f(z)=\frac{(z-2i)(i-4)}{\frac{1-i}{2}z+i+1}$

Now I am almost 100% if the circles would both have center in point 0. However, they have different origins and I somehow don't know how to deal with this kind of problems. :(

I think I would work backwards, and map the parallel lines \mathrm{Re}(z) = 0 and \mathrm{Re}(z) = 1 to the circles |z + 1| = 1 and |z| = 2 respectively by a Mobius map g (these parallel lines are cunningly chosen so that I can take g(0) = 0 and g(1) = 2). I would then need to check that <br /> |g(iy) + 1| = 1 and <br /> |g(1+ iy)| = 2 for all y \in \mathbb{R} (which seems easier than checking that f(z) lies on a particular straight line when |z + 1| = 1 and so forth) and that if \mathrm{Re}(z) \in (0,1) then g(z) lies between the two circles.

(Finally, of course, I must compute f = g^{-1} which what I'm actually asked to find.)

 

[skrat]

Hmm, ok, I found out that for $g(0)=0$, $g(1)=2$ and $g(\infty )=-2$ the Mobius transformation is $g(z)=\frac{z}{1-\frac{1}{2}z}$

Now I don't understand how you came up with following conditions:

$|g(iy) + 1| = 1$ and $|g(1+ iy)| = 2$

For example, the first one:
$|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}|=|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}|=1$

$|\frac{(1+\frac{1}{2}iy)^2}{1-\frac{1}{4}y^2}|=1$

$|(1+\frac{1}{2}iy)^2|=1-\frac{1}{4}y^2$

$2iy=0$

Now it is possible that I made a mistake.. ?

 

[pasmith]

Hmm, ok, I found out that for $g(0)=0$, $g(1)=2$ and $g(\infty )=-2$ the Mobius transformation is $g(z)=\frac{z}{1-\frac{1}{2}z}$

Now I don't understand how you came up with following conditions:

$|g(iy) + 1| = 1$ and $|g(1+ iy)| = 2$

We need the image of \mathrm{Re}(z) = 0 to be |z + 1| = 1. If \mathrm{Re}(z) = 0 then z = iy for some y \in \mathbb{R}. Hence we need |g(iy) + 1| = 1 for all y \in \mathbb{R}.

For example, the first one:
$|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}|=|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}|=1$

$|\frac{(1+\frac{1}{2}iy)^2}{1-\frac{1}{4}y^2}|=1$

You are checking whether the condition |g(iy) + 1| = 1 is true for all y \in \mathbb{R}, so you can't assume this in your calculations. Thus you should start with <br /> |g(iy) + 1| = \left|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}\right|=\left|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}\right| =<br /> \left|\frac{(1 + \frac12 iy)^2}{1 + \frac14y^2}\right| = \frac1{1 + \frac14y^2} \left|(1 + \tfrac12 iy)^2\right|<br />
and you need to show that this is equal to 1. You have now to calculate \left|(1 + \tfrac12 iy)^2\right| = \left|1 - \tfrac14y^2 + iy\right|<br /> = \sqrt{(1 - \tfrac14y^2)^2 + y^2}.<br /> Fortunately, the expression under the square root turns out to be a perfect square.

 

[skrat]

Aaa ok I see it now. We are trying to prove that both of the lines map where they are supposed to; vertical line where $Re(z)=0$ into $|z+1|=1$ and similar for the line $Re(z)=1$.

Second condition:

$|g(1+iy)|=2$

$|\frac{1+iy}{1-\frac{1}{2}(1+iy)}|=2$

After some calculation, we find out that: $|\frac{1+iy}{1-\frac{1}{2}(1+iy)}|=\frac{4}{2}\frac{1+y^2}{1+y^2}=2$

Second condition is also good. Therefore yes, the area between two parallel lines is mapped to the area between the desired circles with this Mobius transformation: $g(z)=\frac{z}{1-\frac{1}{2}z}$

Of course, I need inverse of that, which is $f(z)=\frac{z}{\frac{1}{2}z+1}$

pasmith, thank you very much!