<s>[INST] Domain of analyticity for $\operatorname{Log}(z^4 - 1)^{1/2}$ How do I solve for the analyticity of $\operatorname{Log}(z^4 - 1)^{1/2}$?
So I factored out $$z^4 - 1 = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4 - 1$$
Now the $$\operatorname{Re}(z^4 - 1) = x^4 - 6x^2y^2 + y^4 - 1 < 0 \implies x^4 - 6x^2y^2 + y^4 < 1$$ but I'm stuck here on how to find the set where $\operatorname{Re}(z^4 - 1)$ is nonnegative.
$$\operatorname{Im}(z^4 - 1) = 4x^3y - 4xy^3 = 4xy(x^2 - y^2) = 0 \implies x=0, y=0$$ but unsure how to interpret that either, is it only $z=0$ where it is not analytic or on $x=0$ and $y=0$ axis?
If anyone knows a textbook reference that clearly explains these concepts that would be great as well.
 [/INST] You're doing fairly well. As for the inequality, find that polynomial's roots first:
$$
x^4-6x^2y^2+y^4-1 = 0 \\
x^2 = 3y^2 \pm \sqrt{9y^4-y^4+1} = 3y^2 \pm \sqrt{8y^4+1} \\
x_{1,2,3,4} = \pm \sqrt { 3y^2 \pm \sqrt{8y^4+1}}
$$
So inequality can be rearranged as
$$
(x-x_1)(x-x_2)(x-x_3)(x-x_4) > 0
$$
As for the second region with $\text{Im} (z^4-1) \ne 0$ you didn't find all roots of that equation
$$
4x^3y-4xy^3=0 \\
4xy(x^2-y^2) = 0 \\
\left [ \begin{array}{l}
xy = 0 \\
x^2-y^2 =0
\end{array}
\right . \Longrightarrow \left [ \begin{array}{l}
\left [ \begin{array}{l}
x = 0, y \in \mathbb R \\
y = 0, x \in \mathbb R
\end{array}\right . \\
|x| = |y|
\end{array}
\right .
$$
So final regions looks like filled region below (white color means excluded points)

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