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I have a conformal map that I've been having problems with.

Map the set $\{ z=x+iy : |x|\geq y \}$ with a branch cut on the negative imaginary axis from $[-i,0]$ to the unit disc.

I've tried a few things but the branch cut keeps messing me up.

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2 Answers

up vote 2 down vote accepted

The interior of your domain consists of all points $z=x+iy$ which satisfies one of the three conditions: (i) $x>0$ and $y<x$, (ii) $x=0$ and $y<-1$, or (iii) $x<0$ and $y<-x$.

First map $z$ to $w=iz$. This will rotate your picture counterclockwise by $90^\circ$. Then map $w$ to $\tau=w^{4/3}$. This will close the open sector, and your domain will be mapped to the complex plane minus a cut from $\tau=1$ along the real axis to $\tau=-\infty$. Then map $\tau$ to $\zeta=\sqrt{\tau-1}$, which will send your domain to the right half plane $\Re(\zeta)>0$. Finally map $\zeta$ to ${1-\zeta \over 1+\zeta}$, which will map your domain to the unit disc.

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The region has two rays emanating from the origin as its boundary. One of the rays is in the $i+i$ direction, and the other is in the $-1+i$ direction. This boundary needs to become the boundary of the disc. The region under discussion is the region below these rays. We will do what we need to do to "straighten" out the rays into a vertical line, so that exponentiation will turn them into the boundary of the disc.

Multiply $z$ by $i$, so that the boundary rays are now in the $-1\pm i$ directions. Raise $z$ to the $2/3$ power, using the negative real axis as a branch cut for cube root. Now the boundary rays are the imaginary axis, and the region has nonnegative real parts. Negate $z$, so that the region has nonpositive imaginary parts. And lastly, exponentiate to get the unit disc with its interior.

So all together,

$$z\mapsto \exp(-(iz)^{2/3})$$

where the branch cut for the cube root is the negative real axis. Now, this makes the branch cut for the entire map the positive imaginary axis, which is the opposite of what you asked for. Can you modify this approach?

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What about the "branch [?] cut on the negative imaginary axis from $[-i,0]$"? I think the two sides of this cut also want to make it onto the boundary of the disc. –  Christian Blatter Aug 19 '12 at 19:00
    
@ChristianBlatter A branch cut will put the closed edge on one or of the sides, and leave an open edge on the other side. After the transformation, this ripped seam can glue back together inside the disc, say along a radius. I'll add the full answer now, I just wanted to give OP a chance to think about it more. –  alex.jordan Aug 19 '12 at 20:47
    
@ChristianBlatter Tougher than I thought! I take it back. –  alex.jordan Aug 19 '12 at 21:28
    
Seeing Per Manne's answer, I think I misunderstood the nature of the desired branch cut. It's just supposed to be a slit, not an entire ray. –  alex.jordan Aug 19 '12 at 21:33
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