Conformal mapping. Find the conformal mapping which maps $\{ z \in \mathbb{C}: 0< \operatorname{Im}z<1 \} \backslash[a,a+ih]$ to $\{ z \in \mathbb{C}: 0< \operatorname{Im}z<1 \} $. We assume $a \in \mathbb{R}$ and $0<h<1$.
It seems odd that we can actually "remove" a line segment by conformal mapping. How to proceed exactly? Help needed.
 A: *

*Let $f_1$ be a composition of translation, rotation, and scaling, such that
$$
f_1:\left\{z\in\mathbb{C}:\Im\left(z\right)\in\left(0,1\right)\right\}\setminus\left[a,a+ih\right]\to\left\{z\in\mathbb{C}:\Re\left(z\right)\in\left(0,\pi\right)\right\}\setminus\left[\pi\left(1-h\right),\pi\right].
$$


Recall the mapping $f(z)=-\cos z$. You may check that
$$
f:\left\{z\in\mathbb{C}:\Re\left(z\right)\in\left(0,\pi\right)\right\}\to\mathbb{C}\setminus\left(\left(-\infty,-1\right]\cup\left[1,\infty\right)\right).
$$
Specifically,
$$
f:\left(0,\pi\right)\to\left(-1,1\right).
$$
This is a key step of our method here.


*Consider $f_2(z)=-\cos z$. We have
$$
f_2:\left\{z\in\mathbb{C}:\Re\left(z\right)\in\left(0,\pi\right)\right\}\setminus\left[\pi\left(1-h\right),\pi\right]\to\mathbb{C}\setminus\left(\left(-\infty,-1\right]\cup\left[-\cos\pi\left(1-h\right),\infty\right)\right)
$$

*Now, let $f_3$ be a composition of translation and scaling, such that
$$
f_3:\mathbb{C}\setminus\left(\left(-\infty,-1\right]\cup\left[-\cos\pi\left(1-h\right),\infty\right)\right)\to\mathbb{C}\setminus\left(\left(-\infty,-1\right]\cup\left[1,\infty\right)\right).
$$

*Thereafter, we have
$$
f_2^{-1}:\mathbb{C}\setminus\left(\left(-\infty,-1\right]\cup\left[1,\infty\right)\right)\to\left\{z\in\mathbb{C}:\Re\left(z\right)\in\left(0,\pi\right)\right\}.
$$

*Finally, let $f_4$ be a composition of rotation and scaling, such that
$$
f_4:\left\{z\in\mathbb{C}:\Re\left(z\right)\in\left(0,\pi\right)\right\}\to\left\{z\in\mathbb{C}:\Im\left(z\right)\in\left(0,1\right)\right\}.
$$
To sum up,
$$
f=f_4\circ f_2^{-1}\circ f_3\circ f_2\circ f_1
$$
is the very conformal mapping that meets your demand.
