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For the analyticity of complex $\operatorname{Log}(1-z^{-4})$, I know it is not analytic at branch points $i, -i, 1, -1, 0$ but how does that help determine where the function is analytic?

Thanks.

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    $\begingroup$ Here's a (somewhat off-topic) tip: when you want a subscript or superscript that is more than a single character, enclose it in {} braces. For example, to write $x^{23}$ or $z^{-4}$ write x^{23} or z^{-4}. $\endgroup$ – Aeolian Feb 26 '13 at 1:38
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I assume this is the principal branch of $\log$.

Your map, by composition, is defined and analytic on the set: $$\{z\neq 0\;;\;1-z^{-4}\in \mathbb{C}\setminus (-\infty,0]\}.$$

Let's find out what the complement of the latter is.

It is the set of all $z$ such that $z=0$ or $$ 1-\frac{1}{z^4}=-r $$ for some $r\geq 0$.

This is equivalent to $z=0$ or $$ z^4=\frac{1}{1+r}\quad\Leftrightarrow\quad z=\frac{1}{\sqrt[4]{1+r}} e^{\frac{ik\pi}{2}}. $$

Since $r$ ranges over $[0,+\infty)$ and $z=0$ is included, we see that the complement of the domain is $$ [0,1]1\cup[0,1]e^{i\pi/2}\cup[0,1]e^{i\pi}\cup[0,1]e^{3i\pi/2} $$ which looks like a + centered at $0$.

Your domain of analyticity is the complement of this cross.

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  • $\begingroup$ YES i know thats the definition of where the branch cut is but how do I find that set?!??!?! $\endgroup$ – DJ_ Feb 26 '13 at 6:48
  • $\begingroup$ Ok, ok... No need to yell or use that many question marks... A "could you please be more explicit" would have been way enough. See my edit nevertheless. $\endgroup$ – Julien Feb 26 '13 at 12:05
  • $\begingroup$ Thanks that definitely helped lots ! $\endgroup$ – DJ_ Feb 26 '13 at 17:22

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