Mapping the upper half plane conformally onto a semi-infinite strip,

Map the upper half y>0 of the z-plane conformally onto the semi-infinite strip u>0, $-\pi<v<\pi$ in the w-plane.

I would like some hints for now, please.

I'm not sure how to even get started on this problem, to be honest.

Thanks,

• Think about logarithms... – David C. Ullrich Jun 27 '15 at 17:26
• Hi @DavidC.Ullrich, all I can say, when I think about the complex logarithm mapping, is that the mapping fixes the real line - {0}, which is obvious, but it also maps the upper half plane to the upper half plane - I tested a few points from $C^+$ such as i, 1+i, and -1+i. How does this help? What can I do next? Thanks, – User001 Jun 29 '15 at 23:08
• Actually...it maps the upper half plane to the first quadrant... – User001 Jun 29 '15 at 23:26
• Hmm....I'm testing more points and now notice that the complex log, say, choosing the principal branch, maps all of C - $R^-$ U {0} to the first quadrant. So now I've mapped the upper half plane to Quadrant I. – User001 Jun 29 '15 at 23:39
• You can do better than test sample point. It's easy to show the logarithm maps the upper half-plane to a strip. If $z$ is in the upper half plane what does that say about $z$ in plar coordinates? – David C. Ullrich Jun 30 '15 at 0:37

Let $H$ be the half plane $\{z \in \mathbb{C} \ | \ \Im z > 0 \}$ and : \begin{align} \cosh : \mathbb{C} & \rightarrow \mathbb{C} \\ z & \mapsto \frac{\exp(z) + \exp(-z)}{2}. \end{align}

We have : $$\forall z = x + iy \in \mathbb{C}, \cosh z = \cosh x \cos y + i \sinh x \sin y.\qquad(1)$$

Thus, the preimage of $H$ under $\cosh$ is an union of semi-infinite strips : $$\cosh^{-1} (H) = (\bigcup_{n \in 2 \mathbb{Z}} \{z \in \mathbb{C} \ | \ \Re z > 0 ; n \pi < \Im z < (n + 1) \pi\}) \cup (\bigcup_{n \in 2 \mathbb{Z}} \{z \in \mathbb{C} \ | \ \Re z < 0 ; (n - 1) \pi < \Im z < n \pi\})$$

Let $S$ be the semi-infinite strip $\{z \in \mathbb{C} \ | \ \Re z > 0 ; 0 < \Im z < \pi\}$. I'll show that the restriction $\cosh_{|S}$ of $\cosh$ to $S$ is a biholomorphism onto $H$, from what your problem is easy to solve.

First, let's show that $\cosh_{|S}$ is injective. Let $z, z' \in S$ such that $\cosh z = \cosh z'$. We have $\cosh^2 z = \cosh^2 z'$, so $\sinh^2 z = \sinh^2 z'$, and finally $\sinh z = \pm \sinh z'$. If $\sinh z = \sinh z'$, since $\cosh z = \cosh z'$, we have $\exp z = \exp z'$, and since $z \in S$ and $z' \in S$, this implies $z = z'$. If $\sinh z = - \sinh z'$, we have $\exp z = \exp (-z')$, which is impossible since $\Re z > 0$ and $\Re z' > 0$. Thus $\cosh_{|S}$ is injective.

Next, let's show that $\cosh_{|S}$ is surjective. First, being a holomorphic function, its image is open. Then, thanks to the formula (1), $\cosh$ sends the boundary of $S$ onto the boundary of $H$, and tends towards infinity as its argument grows inside $S$. Thus $\cosh_{|S}$ is a proper function, which implies that its image is closed. Since $H$ is connected, any subset of $H$ which is non-empty, open and closed must equal $H$. Finally, $\cosh_{|S}$ is surjective, and the proof is finished.

• But not conformal (doubt if it possible) – Willemien Feb 8 '17 at 13:46
• First, the map above is clearly conformal since it is a biholomorphism. Then, we know a priori that the problem has a solution thanks to the Riemann mapping theorem, because a semi-infinite strip is a non-empty simply connected open subset of C which is not all C, and so is also the upper half plane, thus both are biholomorphic to the unit disk. – aew Feb 9 '17 at 14:59