analytic function $f$ such that $f(z)^2 = z$ Delete from the complex plane the non-positive part of the imaginary axis. How do we explicitly define an analytic function $f$ on our "modified complex plane" satisfying $f(z)^2 =z$?
This was an exercises from years ago. My old notes just put $f(z)= \sqrt{x}$, which I think is wrong. Also, I have no idea what our professor meant by explicitly. My guess is to show its analyticity, but I'm at a loss on where to start here.
 A: Any $ z \in \mathbb{C} \setminus \mathbb{R}_{-} $ can be written uniquely as $ z = r e^{i \theta} $, where $ r > 0 $ and $ \theta \in (- \pi,\pi) $. Next, define a function $ f: \mathbb{C} \setminus \mathbb{R}_{-} \to \mathbb{C} $ as follows:
\begin{align}
\forall (r,\theta) \in \mathbb{R}_{+} \times (- \pi,\pi): \quad
f(r e^{i \theta}) &= \sqrt{r} e^{i \theta/2} \\
                  &= \sqrt{r} \cos \left( \frac{\theta}{2} \right) + i \left[ \sqrt{r} \sin \left( \frac{\theta}{2} \right) \right] \\
                  &= u(r,\theta) + i \cdot v(r,\theta).
\end{align}
Now, verify that the functions $ u $ and $ v $ satisfy the polar-coordinate version of the Cauchy-Riemann Equations:
$$
\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text{and} \quad \frac{\partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial \theta}.
$$
Once this is done, apply Goursat’s Theorem to deduce that $ f $ is holomorphic on $ \mathbb{C} \setminus \mathbb{R}_{-} $. Finally, your problem is solved upon observing that $ [f(z)]^{2} = z $ for all $ z \in \mathbb{C} \setminus \mathbb{R}_{-} $.
I hope that this is what you are looking for! :)
