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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.

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Take a look at the definition of the principle branch of the square root function first: en.wikipedia.org/wiki/… –  leshik Jan 20 '13 at 5:13
Usually, "explicitly" is used in this context to mean "stating where each $z$ goes in terms of sufficiently elementary functions". Used by people to avoid wise guys who might answer "Well, it's just the analytic function $f:\mathbb{C} - \mathbb{R}^- \to \mathbb{C}$ satisfying $f(z)^2=z$". –  Eric Stucky Jan 20 '13 at 6:54
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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! :)

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I now get it. Thanks! A little nitpicking though: Pardon my ignorance, but doesn't $f$ as defined here give values at the non-positive part of the imaginary axis? –  user58191 Jan 20 '13 at 9:08
Ah! Thanks for pointing that out. It was an oversight on my part. The $ f $ that I constructed is not defined on the non-negative part of the real axis. Anyway, the argument is almost the same. Just replace $ (- \pi,\pi) $ by $ \left( - \dfrac{\pi}{2},\dfrac{3 \pi}{2} \right) $. –  Haskell Curry Jan 20 '13 at 9:19
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