Show that $F$ is an holomorphic function function on $\Omega\smallsetminus\mathbb{R}$ Let be $\Omega = \{z \in \mathbb{C} , z \neq 0, z \neq 1 \}$ 
For $z \in \Omega\smallsetminus\mathbb{R^-}$ we define:
$$F(z) = \sum_{n=-\infty}^{+\infty} \frac{1}{(\log(z) + 2in\pi)^2}$$
where log is the principal determination of logarithme.
I want to Show that $F$ is an holomorphic function function on $\Omega \smallsetminus \mathbb{R^-}$
 A: Let $U = \mathbb C \setminus (-\infty,0].$ For $n=1,2, \dots $ and $z\in U,$
$$|\log z + i2n\pi|^2 = |\ln |z| + i(\text { arg }(z) +2n\pi)|^2 = \ln^2 |z| + (\text { arg }(z) +2n\pi)^2.$$
Since $\text { arg }(z) > -\pi,$ the last term is at least $(2n-1)^2\pi^2.$ Therefore
$$\left | \frac{1}{(\log z + i2n\pi)^2}\right | \le \frac{1}{(2n-1)^2\pi^2}.$$
Thus $\sum_{n=1}^{\infty} \dfrac{1}{(\log z + i2n\pi)^2}$ converges uniformly on $U$ by the Weierstrass M-test. Since a uniformly convergent series of holomorphic functions is holomorphic, we see this series defines a holomorphic function on $U.$ Same for $\sum_{n<0} \dfrac{1}{(\log z + i2n\pi)^2}.$ Thus
$$\sum_{n\in \mathbb Z,n\ne 0}^{\infty} \frac{1}{(\log z + i2n\pi)^2}$$
is holomorphic on $U.$ (If you allow $n=0$ then $F$ has a pole of order $2$ at $1\in U.$) 
A: For $n \in \mathbb N$ and $z \in D$ (where $D=\Omega\setminus \mathbb R_-$), let $$f_n(z)=\frac{1}{(\log(z) + 2in\pi)^2}.$$
$f_n$ is holoporphic and
$$f_n^\prime(z)=-\frac{2}{z(\log z+2in\pi)^3}.$$
For $z_0 \in D$ and using the properties of the complex logarithm, try to define a compact set $K$ such that $z_0$ belongs to the interior of $K$ and such that there exists $A,B > 0$ with $\vert z \vert > A$ and $\vert \log z \vert  >B$ for $z \in K$.
You then have $$\vert f_n^\prime(z) \vert < \left\vert \frac{2}{A(2n\pi-B)^3} \right\vert$$ for $z \in K$. As the sum on $n \in \mathbb N$ of the RHS of above inequality converges, $\sum f_n^\prime(z)$ converges normally on the interior of $K$. This proves that $F$ is holomorphic on a neighborhood of $z_0$. As this this is true for any $z_0 \in D$, we get the desired conclusion.
