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We are working with real integrals and the complex residue theorem. In order to solve the following integral:

$$ \int_0 ^\infty \frac{\sqrt{x}}{1+x^2},$$

I will have to think about how a square root continues "along the complex plane". I am a bit confused about the hint I have gotten, it confuses me more than that it actually helps me. The problem is as follows:

Prove that the function $x \mapsto \sqrt{x}$, defined on $(0,\infty)\to \mathbb R$ has a unique continuation to the analytic function $w: \mathbb C \setminus (-\infty, 0)i \to \mathbb C$

The hint is as follows: use a suitable branch of the logarithmic function. Observe that the notation means we leave out the negative imaginary axis.

We have covered some identities in class with analytic continuation, often involving identities with sines and cosines or exponential laws. The approach was then to show that some function was zero on $\mathbb C$ (using Liouville) and the identity emerged for $\mathbb C$. All the material involving this topic that I have found online talks about branch cuts and about the angles that are then permitted. I am not quite sure how to put it together formally. Essentially we want to define for $z=re^{\phi i}$ and $\phi \in (-\pi/2, 3\pi/2)$: $$ w(z)=\sqrt{r} e^{\frac{\phi i}2}$$ Where do logarithms come in?

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    $\begingroup$ Logarithms are deeply connected to the "argument" (phase factor) of complex numbers. When you write $w(z) = \sqrt{r} e^{i \phi}$, you are effectively saying that $\phi = \text{Im}\left( \log w(z) \right)$. Choosing phase angles to be between $-\pi/2$ and $3 \pi / 2$ is essentially equivalent to choosing to define the logarithm with a branch cut along the negative imaginary axis. $\endgroup$
    – sasquires
    Commented Dec 17, 2020 at 1:54
  • $\begingroup$ Your comment was helpful, I found the section that talks about the logarithm and it indeed defines powers of a complex number, it did not mention that such powers need not be integers, that is probably why I missed it. $\endgroup$
    – user459879
    Commented Dec 17, 2020 at 21:47

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In my lecture notes I found the following definition of the power of a complex number. First we choose a suitable branch of the logarithm, in our case we look at the interval $(\pi/2 - \pi, \pi/2+\pi)$,then we define: $$ z^{\frac{1}{2}}:=e^{\frac{1}{2} \log_{\pi/2}(z)}.$$ Where $\log_{\pi/2}$ has a branch cut along $\{ R e^{\pi/2} | R\in \mathbb R_{\geq 0}\}$.

After this it is indeed shown that such a function is complex differentiable on its domain (which is open, hence this function is holomorphic, hence analytic by Goursat)

The comment by sasquires thus answered my question and I wish to post this so I can close my question as solved.

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