Conform mapping - trying to learn

**liquidFuzz** · September 16th 2010, 12:08 PM

Lets say I have the right half of the real plane. Furthermore the inside of the unit circle. Now I want do a conform mapping of this onto the unit circle. Hmm... I'll try to express my self a bit more mathematically.

$A = \displaystyle \{ z \in \mathbb{C}|~ |z| < 1; ~Re (z) > 0 \}$
$B = \displaystyle \{ z \in \mathbb{C}|~ |z| < 1 \}$
So a conform mapping of A onto B.

How does the proper way of writing this look, or better yet, the right way of doing it. As of now I'd just go for something like. $w = z^2$

**Ackbeet** · September 16th 2010, 12:14 PM

Your mapping will work generally. I'd worry about the boundary points/unusual points like the origin and such. In set A, for example, do you mean that the real part is strictly greater than zero, or is it greater than or equal to zero? Your candidate function is going to leave parts of B out depending on how that all works out.

**Defunkt** · September 16th 2010, 12:15 PM

Every $z \in A$ can be written as $z = e^{i \theta}, \ 0 \le \theta \le \pi$
Every $z \in B$ can be written as $z = e^{i \phi}, \ 0 \le \phi \le 2 \pi$

So if you double the angle of every $z \in A$ , you get every $z \in B$ . $w = z^2$ does just that.

Edit: Didn't notice strict inequality. Thanks ackbeet for pointing that out.

**Ackbeet** · September 16th 2010, 12:31 PM

Technically, you have

$\displaystyle{A=\left\{\rho\,e^{i\theta}:0<\rho<1\ land -\frac{\pi}{2}<\theta<\frac{\pi}{2}\right\}}.$

Also,

$B=\{\rho\,e^{i\theta}:|\rho|<1\}.$

I'm not so sure the mapping $w=z^{2}$ is going to do everything you need. You might need to look into fractional linear transformations.

**Defunkt** · September 16th 2010, 12:43 PM

Oh, wow. I totally read everything wrong.
Just ignore my post :P

**Ackbeet** · September 16th 2010, 12:48 PM

I totally read everything wrong.

Been there, done that. For some reason, this happens especially with logic problems. I love logic, and I've done great in the coursework, but for some reason (probably being over-eager) I tend to mess up when I help people on the MHF. Oh, well.

**liquidFuzz** · September 16th 2010, 01:04 PM

I didn't think about that... But what you say about the $Re > 0$ seems to be right. I should have written $Re \geq 0$ .

At this point solving the Re > 0 mapping is way out of my league. I don't really know where to start. I'll try to get around to it later on. Thanks for the input.

**Ackbeet** · September 16th 2010, 01:06 PM

If you have the possibility of equality there, I think your original candidate will work fine.

**liquidFuzz** · October 21st 2010, 01:13 PM

I've been playing around with different types of standard mappings and I'm starting to get the hang of it. This Möbius Applet helped me tremendously.

$\displaystyle z_1 = iz_2$ - Turns the half circle up to the positive imaginary plane.
$\displaystyle z_2 = \frac{1+z_3}{1-z_3}$ - To map the half circle on the first quadrant.
$\displaystyle z_3 = z_4^2$ - This maps the first quadrant on the upper plane.
$\displaystyle z_4 = \frac{z_5 - i}{z_5 + i}$ - To map the upper plane on the unit circle.

Now all together.
$\displaystyle w = i \left(\frac{1+(\frac{z-i}{z+i})^2}{1-(\frac{z-i}{z+i})^2} \right)$

After some shoveling back and forth we get.
$\displaystyle w = \frac{z^2 -1}{2z}$

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