Why is the $1$-form ${dz\over z}$ on $\mathbb{C}^*$ closed?

Question

Why is the $1$-form $\displaystyle\frac{dz}{z}$ closed in $\mathbb{C}^*$?

In general, how to compute a complex one form's derivative?

Thank you!

Answer 1

Note that

$$d\left(\frac{dz}{z}\right) = d\left(\frac{1}{z}dz\right) = \frac{\partial}{\partial z}\left(\frac{1}{z}\right)dz\wedge dz + \frac{\partial}{\partial\bar{z}}\left(\frac{1}{z}\right) d\bar{z}\wedge dz = 0$$

as $dz\wedge dz = 0$ and $\dfrac{1}{z}$ is holomorphic. Therefore $\dfrac{dz}{z}$ is closed.

In general, a complex one-form on an open subset of $\mathbb{C}$ is of the form $\alpha = f dz + g d\bar{z}$. Then

$$d\alpha = \frac{\partial f}{\partial z}dz\wedge dz + \frac{\partial f}{\partial\bar{z}}d\bar{z}\wedge dz + \frac{\partial g}{\partial z} dz\wedge d\bar{z} + \frac{\partial g}{\partial\bar{z}}d\bar{z}\wedge d\bar{z} = \left(\frac{\partial g}{\partial z} - \frac{\partial f}{\partial\bar{z}}\right) dz\wedge d\bar{z}.$$

A complex one-form $\alpha$ on an open subset of $\mathbb{C}$ is called a holomorphic one-form if $\alpha = f dz$ with $f$ holomorphic. The above computation shows that holomorphic one-forms are closed. Your question is an example of this result as the one-form $\frac{1}{z}dz$ is a holomorphic one-form on $\mathbb{C}^*$.