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 advanced.


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.

  • $\begingroup$ Ah yes, makes sense ! I should have been more careful. Thank you @Antonio Vargas. $\endgroup$ – Gustavo Louis G. Montańo Oct 11 '13 at 14:21
  • $\begingroup$ Glad to help.${}$ $\endgroup$ – Antonio Vargas Oct 11 '13 at 14:30

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