### Differential forms and holomorphic functions

Due by Mon, 02 Apr 2018

We identify $\mathbb{C} = \mathbb{R}^2$ as usual. Let $U \subset \mathbb{C}$ be an open set and let $f: U \to \mathbb{C}$, $f(z) = u(z) + iv(z)$ be a smooth function with $u(z)$ and $v(z)$ the real and imaginary parts of $f(z)$. Let $dz = dx + idy$ and let $f(z)dz$ be given by complex multiplication as usual.

We say that $f(z)$ is holomorphic if $f(z)dz$ is a closed $1$-form, i.e., the real and imaginary parts of $d(f(z)dz)$ are both $0$.

1. Let $\frac{\partial f}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial f}{\partial x} + i\frac{\partial f}{\partial y}\right)$. Show that $f(z)$ is holomorphic if and only if $\frac{\partial f}{\partial \bar{z}} = 0$.
2. Show that if $f(z)$ and $g(z)$ are holomorphic functions on $U$, then $f(z) + g(z)$ is holomorphic.
3. Show that if $f(z)$ and $g(z)$ are holomorphic functions on $U$, then $f(z)g(z)$ is holomorphic.
4. Show that every polynomial $p(z)$ over complex numbers is holomorphic on $\mathbb{C}$.
5. Show that if $f(z)$ is holomorphic on $\mathbb{C}$ and $S^1$ is the unit circle, $\int_{S^1} f(z)dz = 0$.
6. Show that $f(z) = \frac{1}{z}$ is a holomorphic function on $\mathbb{C} \setminus \{0\}$, but $\int_{S^1} f(z)dz \neq 0$.