Injectivity of the function $i\sqrt{2}\sin{z/2}$

Question

Let $U=\{z\in\mathbb{C}:|Re(z)|<\pi\}$. Then why is the function $f:U\rightarrow f(U), f(z)=i\sqrt{2}\sin{z/2}$ injective and is there a way to describe $f(U)$?

Answer 1

A straight-forward calculation shows that $$ \frac{\left( \dfrac{w+i}{w-i} \right)^2 + 1}{\left( \dfrac{w+i}{w-i} \right)^2 - 1} = \frac{1}{2i} \bigl( w - \frac 1w \bigr ) \, . $$ With $w = e^{i z/2}$ it follows that $$ \sin \frac z2 = \frac{\left( \dfrac{e^{i z/2}+i}{e^{i z/2}-i} \right)^2 + 1}{\left( \dfrac{e^{i z/2}+i}{e^{i z/2}-i} \right)^2 - 1} $$ which is a composition of the mappings $$ z \to e^{i z/2}, z \to \frac{z+i}{z-i}, z \to z^2, z \to \frac{z+1}{z-1} $$ Now you can investigate the injectivity and compute the image in four steps:

• Show that $z \to e^{i z/2}$ is injective in $U$ and maps $U$ to the right halfplane.
• Show that $z \to \frac{z+i}{z-i}$ is a bijective mapping of the right halfplane to the upper halfplane.
• Show that $z \to z^2$ is a bijective mapping of the upper halfplane to $\Bbb C \setminus [0, \infty)$.
• Show that $z \to \frac{z+1}{z-1}$ is a bijective mapping of $\Bbb C \setminus [0, \infty)$ to $\Bbb C \setminus ((-\infty, -1] \cup [1, \infty))$.

Finally multiply by $i \sqrt 2$.

In the second and fourth step you can use that Möbius transformations are bijective mappings of the extended complex plane onto itself, and map lines or circles to lines or circles.

Answer 2

Since $$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$

we may write

$$f(z)=\frac{1}{\sqrt{2}}\left(e^{iz/2}-e^{-iz/2}\right)$$

Suppose there's some $z \in U$ and $s \in \mathbb{C}$ with $z+2s \in U$ and $f(z)=f(z+2s)$. Notice the first condition implies $\left| \Re(s) \right|<\pi$, so $s \in U$ too. Then:

\begin{aligned} &e^{iz/2}e^{is}-e^{-iz/2}e^{-is}=e^{iz/2}-e^{-iz/2}\\ \Rightarrow\,\, &e^{iz/2}\left(e^{is}-1\right)=e^{-iz/2}\left(e^{-is}-1\right) \end{aligned}

The terms in parantheses are all nonzero unless $s=2k\pi$ for some $k \in \mathbb{Z}$. Because $s \in U$, this would imply $s=0$ and we'd have nothing to prove. Suppose hence that $s\neq 0$. Then:

\begin{aligned} \Rightarrow\,\, &e^{iz}=\frac{e^{-is}-1}{e^{is}-1}=-e^{-is}=e^{-i(s+\pi)} \end{aligned}

It follows that $z=-(s+\pi)+2k\pi$, or $z+s=(2k-1)\pi$ for some $k\in \mathbb{Z}$. Since $z, s \in U$, the signs of $\Re(z)$, $\Re(s)$ and $(2k-1)\pi$ must all be the same. We conclude that $\left|\Re(z+2s)\right|>\pi$, which contradicts $z+2s \in U$.

Since our assumption that $s\neq 0$ led to a contradiction, we must have $s=0$, that is, $f$ is injective on $U$.

Do you think you can take it from here and describe $f(U)$?