Conformal map entire domain to a strip with specific branchcuts

Question

I am looking for a conformal mapping function that maps the entire z-plane to an infinite strip. (e.g. T=f(z) & -b < Real(T) > b )

I hope to find a function that cuts open to original domain with two branchcuts that lay on one line. (e.g. <-inf,-a] & [a,inf>) such that two point located close to each other but on the oposite side of the branchcut get mapped "far" apart.

Graphical representation:

I tried combining fractional power functions but that did not work out.

Answer 1

Short answer:

$$\frac{2b}{\pi} \arcsin \frac{z}{a}.$$

Verification: Without loss of generality, we assume that $a = 1$ and $b = \frac{\pi}{2}$, the scaling for other values of $a$ and $b$ is trivial to understand. So let's check that the sine maps the strip conformally to the doubly-slit plane:

1. $\epsilon\colon w \mapsto e^{iw}$ maps the strip $\lvert \operatorname{Re} w\rvert < \frac{\pi}{2}$ conformally to the right half-plane. That is well-known.
2. $\varphi\colon \zeta \mapsto \frac{1}{2i}\left(\zeta - \frac{1}{\zeta}\right)$ maps the right half-plane conformally to the doubly-slit plane $\mathbb{C}\setminus \{ t\in\mathbb{R} : \lvert t\rvert \geqslant 1\}$. Solving the quadratic $2i\zeta z = \zeta^2 - 1$ gives $\zeta = iz \pm \sqrt{1-z^2}$, and choosing the branch of the root with $\sqrt{1-0^2} = 1$, we find the solution in the right half-plane as $\zeta = iz +\sqrt{1-z^2}$ while $iz-\sqrt{1-z^2}$ gives the solution in the left half-plane, showing the conformality of $\varphi$.

Longer answer:

Now, if one doesn't know that $\sin$ resp. $\arcsin$ give the desired biholomorphisms, how could one go about finding a conformal map between the doubly-slit plane and the strip? It is, for me at least, easier to map the strip to the doubly-slit plane, so I'll do that. We have two points where we want to close a straight angle, which can be done by squaring.

In the Riemann sphere, squaring closes straight angles at $0$ and at $\infty$, so we first want to map the strip to a half-plane so that the two points where we want to close the straight angle are mapped to $0$ resp. $\infty$.

That we can do in two steps, first map the strip to the right half-plane, and then apply an automorphism that places the two points where we want them. The first part is of course done by $w \mapsto e^{iw}$, which places the two points $\pm\frac{\pi}{2}$ at $\pm i$. Then to position the two points, we want a Möbius transformation that sends $-i \mapsto 0$, $i \mapsto \infty$ and $\infty$ to a point on the imaginary axis in the lower half-plane (so that the right half-plane is mapped to itself), picking $\infty \mapsto -i$ seems to be a good choice. That gives the map

$$T\colon \zeta \mapsto -i\frac{\zeta+i}{\zeta-i},$$

so $w \mapsto T(e^{iw})$ maps the strip conformally to the right half-plane, sending $-\frac{\pi}{2}$ to $0$ and $\frac{\pi}{2}$ to $\infty$. Now $\eta \mapsto \eta^2$ conformally maps the right half-plane to $\mathbb{C}\setminus \{t\in\mathbb{R} : t \leqslant 0\}$, and we can use another Möbius transformation to conformally map that to $\mathbb{C}\setminus \{t\in \mathbb{R} : \lvert t\rvert \geqslant 1\}$. To achieve that, we must map $0 \mapsto-1$, $\infty \mapsto 1$, and we can pick a point on the negative real half-axis to map to $\infty$, $-1\mapsto \infty$ seems like a good choice, giving

$$S\colon u \mapsto \frac{u-1}{u+1}$$

and altogether

$$\begin{align} w &\mapsto S\left(T\left(e^{iw}\right)^2\right)\\ &= \frac{T\left(e^{iw}\right)^2-1}{T\left(e^{iw}\right)^2+1}\\ &= \frac{-\left(\frac{e^{iw}+i}{e^{iw}-i}\right)^2-1}{-\left(\frac{e^{iw}+i}{e^{iw}-i}\right)^2+1}\\ &= \frac{(e^{iw}+i)^2 + (e^{iw}-i)^2}{(e^{iw}+i)^2 - (e^{iw}-i)^2}\\ &= \frac{2(e^{2iw}-1)}{4ie^{iw}}\\ &= \frac{e^{iw}-e^{-iw}}{2i}\\ &= \sin w. \end{align}$$