complex-analysis
How do you find a map from the angle $\{z\in\mathbb{C}:\pi/6<\arg(z)< \pi/2\}$ to the upper half-plane? Is this the same as finding what values $z$ can take in the complex plane, and is just a wedge?
And in general, how do you approach problems like this? I know that if it was instead not to the upper half plane and something like $z\mapsto\frac{1}{z}$, we could just say the map is then $\{z\in\mathbb{C}:0>\arg(z)>-\frac{\pi}{3}\}$ … correct?
Any help is appreciated. Thanks.
Let \begin{align}
\varphi &: z \mapsto \exp\left(\frac{\,z\,}{2}\right),\\
\psi &: z\mapsto i\frac{z-1}{z+1},\\
\phi &: z \mapsto \sqrt{z},\\
g &: z \mapsto \left(\frac{1+z}{1-z}\right)^2,
\end{align}
where $\sqrt{z}$ is defined in $\mathbb{C}\setminus [0,\infty)$ so that $\sqrt{re^{i\theta }}=\sqrt{r}e^{i\theta /2} $ for $0<\theta <2\pi$.
Also we define
\begin{align}
D&=\{z\in \mathbb{C} : |\operatorname{Im}\, z|<\pi\}\setminus [-\pi i; 0],\\
D_1&=\{z\in \mathbb{C} : \operatorname{Re}\, z>0\}\setminus \left\{z : z=e^{i\theta}, -\frac{\pi}{2}<\theta \le 0\right\},\\
D_2&=\{z\in \mathbb{C} : |z|<1\} \setminus [0; 1),\\
D_3&=\{z\in \mathbb{C} :\operatorname{Im}\,z>0, |z|<1\}.
\end{align}
Note that
\begin{align}
&\varphi : D\to D_1,\quad \psi : D_1\to D_2, \quad \phi : D_2\to D_3,\\
&g : D_3\to \mathbb{H},
\end{align}
where $\mathbb{H}$ is the upper half-plane. See the diagram below.
Thus we get $$
f(z)=g\circ\phi\circ\psi\circ\varphi (z)=\left(\frac{1+\sqrt{i\frac{e^\frac{z}{2}-1}{e^\frac{z}{2}+1}}}{1-\sqrt{i\frac{e^\frac{z}{2}-1}{e^\frac{z}{2}+1}}}\right)^2
$$
as a desired mapping function.
Indeed the imaginary axis is mapped to the real axis.
However, the circle $|z|=1$ is mapped to a line (because $i\to \infty$) containing the origin (because $-i\to 0$). This line is orthogonal to the real axis (conformality of mapping at $-i$), and is therefore the imaginary axis.
Your mapping carries the semicircle into a quadrant. Which ? Use conformality. The interval from $-i$ to $0$ is mapped to the interval from $0$ to $-1$. Comformality implies that what was to the right walking along first interval will be to the right walking along its image. This implies that the semicircle will be mapped to the second quadrant.
To get the result you wanted, you need to rotate your function by $-\pi/2$ (to map to first quadrant), and then square (to map to upper half plane).
Therefore the mapping would be
$$-1\frac{(z+i)^2}{(z-i)^2}.$$
Best Answer
You want to stretch out this wedge of (angular) width $\frac\pi2 - \frac\pi6 = \frac\pi3$ to the entire half-plane, of width $\pi$, which is an expansion by a factor of $3$. But you also want to turn about the origin so that the arguments land between $0$ and $\pi$.
To stretch the argument by a factor of $3$, use $z \mapsto z^3$. Then to shift the argument (rotate in the complex plane) back to the upper half-plane, multiply by $e^{-i\pi/2} = -i$. Composing these two operations yields the function $$ f(z) = -iz^3. $$
You can check that for any $z = r e^{i\theta}$ in polar form, $$ f(r e^{i\theta}) = -i\,(r e^{i\theta})^3 = r^3 e^{i(3\theta - \pi/2)}, $$ and that $\theta \mapsto 3\theta - \pi/2$ maps the interval $(\pi/6, \pi/2)$ bijectively onto $(0, \pi)$.
You can generalize this result. To map the wedge $$ D(\alpha, \beta) = \{ z = re^{i\theta} \in \mathbb{C} \mid r>0 \text{ and } \alpha < \theta < \beta \} $$ bijectively onto the upper half-plane $H = D(0, \pi)$, you have to determine the appropriate stretch factor and shift for the arguments, i.e. the appropriate affine linear $\theta \mapsto m \theta + c$ sending the interval $(\alpha, \beta)$ to $(0, \pi)$. The function $$ \theta \mapsto \frac{\pi\, (\theta - \alpha)}{\beta - \alpha} $$ works, provided that $0 < \beta - \alpha < 2\pi$. You can construct the bijective map $D(\alpha, \beta) \to H$ from this.