Why should $f(z)=\sqrt{z}$ be limited on $\mathbb{C}-\{z:\Re(z)\leq0\}$ to be considered as an analytic function?

Question

A multivalued function $f(z)$ can be analytic on an open set $\Omega$ where $f(z)$ has an unique value and is differentiable on every point. If $f(z)=\sqrt{z}$, I think $\Omega$ can be defined as $\mathbb{C}$, instead of $\mathbb{C}-\{z:\Re(z)\leq0\}$, which is the condition usually required for this function.

If $z=r e^{i\theta}$ with $-\pi<\theta\leq \pi$, then $\sqrt{z}=\sqrt{r}e^{i\theta/2}$. Thus, $\sqrt{z}$ is clearly a single-valued function on $\mathbb{C}$.

Could you tell me why $\Omega = \mathbb{C}-\{z:\Re(z)\leq0\}$ instead of $\Omega=\mathbb{C}$? Is it because of continuity or differentiability of $f(z)$ on negative real line?

Answer 1

Your proposed function isn't even continuous on $\{z \in \mathbb{C} \mid \operatorname{Re} < 0, \operatorname{Im} = 0\}$ let alone differentiable. For example, note that

$$\lim\limits_{\theta \to \pi^-}\sqrt{e^{i\theta}} = \lim\limits_{\theta \to \pi^-}e^{i\theta/2} = e^{i\pi/2} = i$$

but

$$\lim\limits_{\theta \to -\pi^+}\sqrt{e^{i\theta}} = \lim\limits_{\theta \to -\pi^+}e^{i\theta/2} = e^{-i\pi/2} = -i.$$

Therefore $\lim\limits_{z \to -1}\sqrt{z}$ doesn't exist, so it isn't continuous or differentiable at $z = -1$.

Answer 2

$$\text{Hint}:\ \cdot f(z) = \sqrt{z} = e^{\frac{1}{2} \ \log(z)}.$$

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