I need some help in determining where $\log(z+z^{-1} -2)$ is analytic, where $z$ is a complex number and $\log(z)=\ln|z|+\arg(z+2k\pi),k\in\mathbb{Z}$. Thank you in advance.
Where is $\log(z+z^{-1} -2)$ analytic?
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Too long for a comment. The Maple 17 commands $$expression := log(z+1/z-2); FunctionAdvisor(branch\_cuts, expression, plot = `2D`) $$ produce

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We have a composition f(g(z)) of functions, with $f(z)=Logz$, and $g(z)=\frac{(z-1)^2}{z}$ so we need g(z) to be defined in the domain of $Logz$, which is $\mathbb C-[0,\infty)$.
Let $w:=\frac{(z-1)^2}{z}$ .We want to exclude $w:=\frac{(z-1)^2}{z}$ from the branch cut $[0,\infty)$. This means we want to exclude the cases where:
1)$Im(w)=0$ and
2)$Re(w) \leq 0$
Notice we use and (instead of or) since we want to exclude points who have both a real part less than $0$ , and an imaginary part equal to $0$.
Expanding on w, letting $z:=(a+ib)$, we get :
3)$w= a^3-2a^2+a-2b^2+ab^2+i(a^2b+b^3-b)$ (I double-checked with Wolfram.)
Notice there is a nice factorization for the imaginary part of $w$;
4)$i(a^2b+b^3-b)=ib(a^2+b^2-1)$.
So, let's check #4 to see where 1),2) are satisfied:
From #4, we see that $Re(w)<0$ if either:
$I)b=0$
$II)a^2+b^2-1=0$
Let's check the case $I$ first , i.e., let's see what happens when $b=0$. If $b=0$,
using the expression for $w$ in #3 , we want to see where we get $b=0$ and $Re(w)<0$;
if $b=0$ , then$Re(w)=a^3-2a^2+a$. We see that $Rew=a(a-1)^2 \leq 0$ only when $a<0$, since
$(a-1)^2$ is always nonnegative. So we have that $Rew \leq 0$ precisely in $(- \infty,0]$
So we need to exclude the set :$Imz=0, Rez \leq 0$ , i.e., we need to exclude the negative real axis.
Now we must do something similar for the case where $a^2+b^2-1=0$ . Notice there is
a nice factorization for $Rew$; $Rew=(a^2+b^2)(a-2)+a$ . This should help you figure out
what part of $Rez$ to exclude for this case where $a^2+b^2-1=0$ . Can you finish it?
$$z+z^{-1}-2=\frac{z^2-2z+1}z=\frac{(z-1)^2}z$$
If you want to "delete" the usual branch cut for the logarithmic function, i.e. the ray $\,(-\infty,0]\;$ on the real axis, then it must be
$$\frac{(z-1)^2}z\notin (-\infty,0]\iff \text{Re}\left(\frac{(z-1)^2}z\right)>0\;\;\vee\;\;\text{Im}\left(\frac{(z-1)^2}z\right)\ne k\pi\;,\;k\in\Bbb Z$$