Holomorphic Functions

The complex upper half plane is the set

$$\mathfrak{h}= \{z\in\mathbb{C}\,:\,$$   Im$$(z) > 0\}.$$

A holomorphic function $f:\mathfrak{h}\rightarrow \mathbb{C}$ is a function such that for all $z\in\mathfrak{h}$ the derivative

$$f'(z) = \lim_{h\rightarrow 0} \frac{f(z+h) - f(z)}{h}$$

exists. Holomorphicity is a very strong condition because $h\in\mathbb{C}$ can approach 0 in many ways.

Example 2.1   Let $\SL_2(\mathbb{Z})$ denote the set of $2\times 2$ integers matrices with determinant $1$. If $\gamma=\left( \begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)\in\SL_2(\mathbb{Z})$, then the corresponding linear fractional transformation

$$\gamma(z) = \frac{az+b}{cz+d}$$

is a holomorphic function on $\mathfrak{h}$. (Note that the only possible pole of $\gamma$ is $-\frac{d}{c}$, which is not an element of  $\mathfrak{h}$.)

For future use, note that if $f:\mathfrak{h}\rightarrow \mathbb{C}$ is a holomorphic function, and $\gamma=\left( \begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)\in\SL_2(\mathbb{Z})$, then

$$f\vert _\gamma(z) = f(\gamma(z)) (cz+d)^{-2}$$

is again a holomorphic function.

Example 2.2   Let $q(z) = e^{2\pi i z}$. Then $q$ is a holomorphic function on  $\mathfrak{h}$ and $q'=2\pi i q$. Moreover, $q$ defines a surjective map from $\mathfrak{h}$ onto the punctured open unit disk $D=\{z\in\mathbb{C}: 0<\vert z\vert<1\}$.