I want show that the function $w: \mathbb{C}^{*} \rightarrow \mathbb{C}$ determine by:
$$ w(z) = \frac{1}{z} $$
Hasn't a primitive function defined in $\mathbb{C}^{*}$, I have a primitive function of $w(z)$ for the set $\mathbb{C}^{-} = { z \in \mathbb{C}-(-\infty, 0]}$, but I don't know how I could proceed.
Thank you.
As Andrew D. Hwang says, this is hard to answer when we don't know what tools are available to you, since you have rejected the obvious choice. But I will assume you at least know that the only primitives of $w(z) = 0$ on a region (which is connected by definition) are constants.
Now consider that primitive you have on $\Bbb C^-$, which we refer to as $\ln z$. Suppose that you have a function $f$ on $\Bbb C^*$ with $f'(z) = 1/z$ for $z \in \Bbb C^-$. Then we can define $g(z) = f(z) - \ln z$ on $\Bbb C^-$ and discover that $g'(z) = 1/z - 1/z = 0$. I.e., $g$ is a primitive of $0$ and is therefore some constant $C$. So $f(z) = \ln z + C$ for some $C$. But $$\lim_{\theta \to \pi} f(e^{i\theta}) = C + i\pi \ne C - i\pi=\lim_{\theta \to -\pi} f(e^{i\theta})$$ while $$\lim_{\theta \to \pi} e^{i\theta} = -1=\lim_{\theta \to -\pi} e^{i\theta}$$ So $f$ cannot be continuous at $-1$, which means it cannot be analytic on $\Bbb C^*$.
