# $f(z)=\frac{1}{z}$ primitive function

In class, my professor showed that the function:

$$f(z)=\frac{1}{z}$$

does not have a primitive function in the domain $$\mathbb C$$ since:

$$\int_{|z|=R}=2\pi i$$

She then said that it does has a primitive function for the domain $$\mathbb C \setminus \left\{x+iy,Im(z)=0, z\le 0\right\}$$, and that function is $$F(z)=log(z)$$.

She did not explain why this is true.

How should I approach this in order to solve this myself?

## 2 Answers

For each $$z\in\mathbb C\setminus(-\infty,0]$$, let $$\gamma_z(t)=1+t(z-1)$$ ($$t\in[0,1]$$). Define$$F(z)=\int_{\gamma_z}\frac1w\,\mathrm dw.$$Then $$\bigl(\forall z\in\mathbb C\setminus(-\infty,0]\bigr):F'(z)=\frac1z$$.

The difference is that in $$\mathbb{C}\setminus (-\infty,0]$$ there is a branch of logarithm. Let's call it $$L$$ and this is by definition of a branch a continuous function. Then at any point $$z_0$$:

$$\lim_{z\to z_0}\frac{L(z)-L(z_0)}{z-z_0}=\{w=L(z)\}=\lim_{w\to w_0}\frac{w-w_0}{e^w-e^{w_0}}=\frac{1}{e^{w_0}}=\frac{1}{z_0}$$

Where I used the known result that $$(e^w)'=e^w$$.