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$\newcommand{\Log}{\operatorname{Log}}$

The question stands as

Where is the function $\Log(z^2-1)$ analytic

, where $\Log$ stands for the principal complex logarithm. My understanding is that

The domain of analyticity of any function $f(z) = \Log\left[g(z)\right]$, 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 $\left \{z = x + iy\ |\ −\infty < x \leq 0, y = 0\right \}$.

Following this definition it would imply that the function $f(z)$ is analytic everywhere in complex plane except for the points where $-\infty<\Re(z^2-1)\leq0$ and $\Im(z^2-1)=0$. So I get $x^2-y^2-1\leq0$ and $2xy=0$. Graphically it must be analytic everywhere except on the real x axis, the imaginary y-axis and in the region inside the hyperbola $x^2-y^2=1$. The answers say

Everywhere except $\{z\in\mathbb{R}:|z|\leq1\}\bigcup\{iy:y\in\mathbb{R}\}$.

Please help correct my understanding. Thank you in advance.

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1 Answer 1

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If $2xy=0$ then either (a) $x=0$, in which case the other inequality becomes $-y^2-1\leq 0$ which is satisfied by all $y\in\mathbb{R}$, or (b) $y=0$, where the other inequality becomes $x^2 - 1 \leq 0$ which is satisfied by all $|x| \leq 1$.

These inequalities must both be satisfied together. You are describing the union of the sets they are satisfied on individually, where what you really want is the intersection.

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  • $\begingroup$ Ah yes, makes sense ! I should have been more careful. Thank you @Antonio Vargas. $\endgroup$ Commented Oct 11, 2013 at 14:21
  • $\begingroup$ Glad to help.${}$ $\endgroup$ Commented Oct 11, 2013 at 14:30
  • $\begingroup$ Can you please explain why there should be intersection at the end and not the union? I don't understand it because (a) and (b) are mutually exclusive. $\endgroup$
    – Koro
    Commented Jun 15, 2023 at 20:24
  • $\begingroup$ @Koro You dont need to think about this question algebraically. The principal branch has its branch cut on the negative real axis. So you are being asked to find the set which $z^2$ maps to $(-\infty, 1]$. The $[0,1]$ part of that is mapped onto by $[-1,1]$ and the remaining $(-\infty,0)$ part is mapped onto by the whole imaginary axis. So this function is analytic outside this set. You take the intersection because the inequalities being referred to are for Re(z^2-1) and Im(z^2-1) respectively. But this is unnecessarily complicated because you can just do it the way I mentioned. $\endgroup$ Commented Jun 16, 2023 at 7:23

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