If $f$ is ananlytic in $D_r(z_0)$\ {$z_0$} and $Ref(z)>0$ for all $z\in D_r(z_0)$\ {$z_0$} then $z_0$ is a removable singularity If $f$ is ananlytic in $D_r(z_0)$\ {$z_0$} and $Ref(z)>0$ for all $z\in D_r(z_0)$\ {$z_0$} then $z_0$ is a removable singularity. I want to prove this statement but I just cannot seem to find a way. Clearly $f(z)\neq0$ in $D_r(z_0)$\ {$z_0$}. I just feel like this can be proved with this fact. Can someone give me a hint? Assistance is appreciated. Thanks a lot
 A: $T(z) = (z-1)/(z+1)$ is a Möbius transformation and maps the
right half plane $\{ \operatorname{Re} z > 0 \}$ to the unit disk.
As a consequence, $g(z) = T(f(z)) = (f(z)-1)/(f(z)+1)$ is analytic and bounded
in $D_r(z_0)\setminus \{z_0\}$. It follows from 
Riemann's theorem
that $g$ has a removable singularity at $z_0$.
We have $g(z_0) \ne 1$ because of the maximum modulus principle,
so that
$f(z) = (1+g(z))/(1-g(z))$ has a removable singularity at $z_0$.
A: Consider $g(z)=e^{-f(z)}.$
Notice that $g(z)$ is bounded on $D(r,z_0)$ then $g(z)$ is holomorphic  by Riemann's theorem, let $h(z)$ some holomorphic logarithm   of $g(z)$ i.e. $g(z)=e^{h(z)}$ then $f(z)=h(z)+2k\pi i$ where $k\in \mathbb{Z}$ ($k$ is constant by continuity and connectedness) therefore $f(z)$ is holomorphic on $D(r,z_0)$.
A: More general result exists:
$f$ is analytic on $B_r(z_0) \setminus \left\{z_0\right\}$  and $Im (f) $ is contained in a simply connected domain $\mathrm{\Omega}$ (where $\mathrm{\Omega} \ne \mathbb{C} $ ). Then $f$ has a removable singularity at $z_0$. 
Proof: This is a corollary of Riemann Mapping Theorem. Since $\mathrm{\Omega}$ is biholomorphic to the unit open disk $\mathbb{D}$, we can assume that $Im (f) \subset \mathbb D$. But then $f$ is bounded and hence the singularity must be removable. QED
To avoid using Riemann Mapping Theorem for this particular problem, you can use the Cayeley Transformation to get hold of the required biholomorphism. To prove the general statement, use Casorati–Weierstrass theorem to Eliminate the essential singularity case. If it were a pole, it would emit rays eventually in all directions,which is a contradiction to the fact that the range is contained in a simply connected domain other than $\mathbb{C}$. 
