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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?

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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 $\bigl(\forall z\in\mathbb C\setminus(-\infty,0]\bigr):F'(z)=\frac1z$.

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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 $(e^w)'=e^w$.

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