# Homework Help: Conformal mapping

1. Apr 29, 2014

### skrat

1. The problem statement, all variables and given/known data
Find function that maps area between $|z|=2$ and $|z+1|=1$ on area between two parallel lines.

2. Relevant equations

3. 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. :(

2. Apr 29, 2014

### pasmith

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.

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

3. Apr 30, 2014

### 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.. ?

4. Apr 30, 2014

### pasmith

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 $y \in \mathbb{R}$.

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 $$|g(iy) + 1| = \left|\frac{iy}{1-\frac{1}{2}iy}+\frac{1-\frac{1}{2}iy}{1-\frac{1}{2}iy}\right|=\left|\frac{1+\frac{1}{2}iy}{1-\frac{1}{2}iy}\right| = \left|\frac{(1 + \frac12 iy)^2}{1 + \frac14y^2}\right| = \frac1{1 + \frac14y^2} \left|(1 + \tfrac12 iy)^2\right|$$
and you need to show that this is equal to 1. You have now to calculate $$\left|(1 + \tfrac12 iy)^2\right| = \left|1 - \tfrac14y^2 + iy\right| = \sqrt{(1 - \tfrac14y^2)^2 + y^2}.$$ Fortunately, the expression under the square root turns out to be a perfect square.

5. Apr 30, 2014

### 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!