I have the Möbius transformation $f(z)=\frac{z-1}{1+z}$ and I am being asked to describe the image of $\{z\in\mathbb{C}|0\leq arg(z)\leq\theta\}$ under $f$, where $\theta\in(0,2\pi)$. I would know how to do this when $\theta$ is not arbitrary (e.g. say $\theta=\pi/2$) by considering the image of the $2$ boundary lines $\theta=0$ and say $\theta=\pi/2$, and then some point in the interior. But I don't know how to do it for some general $\theta$. I assume that $f(0)=-1$ and $f(\infty)=1$ are important to note, since $0,\infty$ are points on any line $arg(z)=\theta$, but to know the image of this line I would need the image of another, third point on it. This is where I am stuck, since considering $f(e^{i\theta})$ seems to get me nowhere.
Image of arbitrary sectors under Möbius transformation
complex numberscomplex-analysismobius-transformation
Best Answer
This covers the case $\theta \in (0, \pi)$, the other cases follow from the same ideas.
You also know that $f(-1) = \infty$, so the image of the real axis is the real axis. Since Moebius transformations send lines to lines or circles, the line $L$ containing the ray $arg(z)= \theta$ must be sent to a line or a circle containing $-1 = f(0)$ and $1 = f(\infty)$. But then $f(L)$ must be a circle $C$, since the only line containing $-1$ and $1$ is the real axis, which is its own image.
Now you only need to find $C$. But you know that Moebius transformations preserve angles, so the angle between $C$ and the real axis at $z = -1 = f(0)$ must be $\theta$ (since this is the angle between the preimages of the real axis and $C$). There is only one circle passing through -1 and 1 that satisfies this condition, so the image of the sector is the region bounded by $C$ and the real axis and contained in the upper half plane. It looks like this:
Using a little trigonometry you can get the center of $C$ (it must be on the imaginary axis, since $-1$ and $1$ are both on $C$) and its radius, if you want explicit inequalities describing the region.