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i would like to find the domain for $f(z)=\sqrt{z^2-a^2},\,a\in\mathbb{R}$ and square root with positive imaginary part. My idea:

$f(z)=\exp\left(\frac{1}{2}\operatorname{Log}(z^2-a^2)\right)$

So $f$ is not analytic for $z=\pm a$ and for all $z$ with $\Im(z)=0$ and $|z|\leq a$, since the domain of analyticity of any function $f(z)=\operatorname{Log}(g(z))$, where $g(z)$ is analytic, will be the set of points $z$ such that $g(z)$ is defined and $g(z)$ does not belong to the set $\{z=x+iy | -\infty<x\leq0,\,y=0\}$.

Am I right?

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The function $$ f(z)=\sqrt{z^2-a^2} $$ is definable as an analytic function in any domain $U\subset\mathbb C$, such that $a$ and $-a$ belong to the same connected component of $\mathbb C\cup\{\infty\}\setminus U$.

The most typical such domain is $U=\mathbb C\setminus[-a,a]$.

This is due to the following lemma:

Lemma. If $a$ and $b$ belong to the same connected component of $\mathbb C\cup\{\infty\}\setminus U$, then $$ F(z)=\log\left(\frac{z-a}{z-b}\right) $$ defines an analytic function is $U$.

Using the Lemma, we may define $$ \sqrt{z^2-a^2}=(z-a)\exp\left(\frac{1}{2}\log\left(\frac{z-a}{z-b}\right)\right) $$

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  • $\begingroup$ Thanks! Solution says it is not analytic for $z=\pm a$ and for all $|z|\geq a$. So that's wrong, because that's not connected?! $\endgroup$ – user411437 Dec 21 '17 at 11:02

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