Showing that $a$ is a removable singularity if $\mathrm{Im}(f(z))$ is bounded from above 
Problem: Suppose $f$ is analytic on the domain $\Omega$ except at the isolated
  singularity $a \in \Omega$.  Show that $a$ is a removable singularity
  if $\mathrm{Im}(f(z))$ is bounded from above.

Attempt:


*

*We have that $a$ is a removable singularity if and only if we have that
$$
\lim_{z\rightarrow a} (z-a)f(z) =0
$$

*Since $\mathrm{Im}(f(z))$ is bounded from above, there exists some $M \in \mathbb{R}_{\ge 0}$ s.t. $\mathrm{Im}(f(z)) \le M$ for all $z \in \Omega$.

*If we could show 
$$
\mathrm{Im}(f(z)) \le M \implies |f(z)| \le M_1 \text{ for some } M_1 \in \mathbb{R}_{\ge 0}
$$
then we would immediately have (1) since 
$$
\lim_{z \rightarrow a}|(z-a)f(z)| \le \lim_{z \rightarrow a}|(z-a)|M_1 = 0
$$
Question: Is this the right approach?
 A: The function $g:=-if$ satisfies $Re(g)=Im(f)$, so $Re(g)$ is bounded above near $a$; and of course it is enough to show that $g$ has a removable singularity at $a$.
Choose $r>0$ such that $Re(g)$ is bouded above on $D^*(a,r):=D(a,r)\setminus\{ a\}$. Consider the function $\phi=e^g$. This function is holomorphic in $D^*(a,r)$ and bounded therein since $\vert \phi\vert=e^{Re(g)}$. Hence, $\phi$ has a removable singularity at $a$. Let us still denote by $\phi$ the analytic continuation of $\phi$ defined on $D(a,r)$.
Write $\phi(z)=(z-a)^k\psi (z)$, where $k$ is a nonnegative integer and $\psi$ is holomorphic in $D(a,r)$ with $\psi(a)\neq 0$. Then $\psi$ has no zeros in $D(a,r)$ because $\phi=e^g$ has no zeros in $D^*(a,r)$. So one can find a function $h$ holomorphic in $D(a,r)$ such that $e^{h(z)}=\psi(z)$ in $D(a,r)$. Then $\phi(z)=(z-a)^ke^{h(z)}$, so that 
$$\forall z\in D^*(a,r)\;:\; e^{g(z)}=(z-a)^k e^{h(z)}\, .$$
In particular, the function $u_k(z)= (z-a)^k$ has a holomorphic logarithm on $D^*(a,r)$, namely $l(z)=g(z)-h(z)$. This forces $k=0$. Indeed, if $C\subset D^*(a,r)$ is any (positively oriented) circle with center $a$, then on the one one hand the winding number of $u_k(C)$ around $a$ must be equal to $0$ since $u_k$ has a holomorphic logarithm, and on the other hand this winding number is equal to $k$. "Alternatively", observe that  $$2i\pi k= \int_{\partial D(a,r/2)} \frac{k}{z-a}\, dz=\int_{\partial D(a,r/2)} l'(z)\, dz=0\, .$$
So we obtain that $e^{g(z)-h(z)}\equiv 1$ on $D^*(a,r)$. This means that $g(z)-h(z)\in2i\pi \mathbb Z$ for all $z\in D^*(a,r)$, and since $D^*(a,r)$ is connected, it follows that there exists an integer $n\in\mathbb Z$ such that
$$\forall z\in D^*(a,r)\;:\; g(z)-h(z)=2i\pi n\, .$$
Hence, the function $h+2i\pi n$ is an analytic continuation of $g$ in the disk $D(a,r)$, which gives the required conclusion.
A: Well, as you say, you certainly have $$\begin{eqnarray}
 1. && |f| \text{ bounded} \Rightarrow \lim_{z\to a} (z-a)f(z) = 0 \\
 2. && |f| \text{ bounded} \Rightarrow \textrm{Im }f \text{ bounded.}
\end{eqnarray}$$
To show that $\lim_{z\to a} (z-a)f(z) = 0$ implies $\textrm{Im } f$ is bounded, you'd need the reverse direction of (1), and to shows that $\textrm{Im } f$ is bounded implies $\lim_{z\to a} (z-a)f(z) = 0$ the reverse direction of (2).
Both reverse directions will need to use that $f$ is analytic, because they aren't true otherwise. You can get the reverse direction of (1) by looking the the laurent series of $f(z)$ around $a$, I think. Dunno how you'd get the reverse direction of (2) - though maybe looking at the laurent series of $f$ yields something again, if you choose your $x$ carefully...
A: Holomorphic functions are very very firms. So I thought: there must be a link (beyond Cauchy-Riemann conditions) between the behaviour of the real and the imaginary part of a such function.
Let
$
f:\Omega\setminus\{a\}\longrightarrow\mathbb C
$
be an holomorphic function where $a$ is an isolated singularity and $\exists M\ge0$ s.t. $|v|\le M$ on $\Omega\setminus\{a\}$, with $f=u+iv$.
Then $a$ is a removable singularity.
Let's show that if $a$ is not a removable singularity then $v$ can't be bounded on $\Omega\setminus\{a\}$.
There are only two possibilities for $a$ to be not a removable singularity: be a pole or an essential singularity.

Case 1: $a$ is a pole.
Let's write the Laurent series for $f$ around $a$, pole of order $m$:
$$
f(z)=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n}+
\sum_{n=0}^{+\infty}c_{n}(z-a)^n\;,\;\;\;z\in B(a,r[\setminus\{a\}
$$
where
$$
P_{f,a}(z)=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n}
$$
is the principal (or sometimes singular) part of $f$ around $a$. The name "principal part" is because the nature of the singularity is its her responsability.
Now, $a$ is a pole for $f$ iff $\lim_{z\rightarrow a}f(z)=\infty_{\mathbb C}$, and naturally this holds iff $\lim_{z\rightarrow a}P_{f,a}(z)=\infty_{\mathbb C}$.
Then, being $v$ bounded, $\lim_{z\rightarrow a}f(z)=\infty_{\mathbb C}$ iff $\lim_{z\rightarrow a}u(z)=\infty$ which implies that $\lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty$. Moreover $v$ bounded implies $\Im P_{f,a}$ bounded.
Let's now reach the contradiction showing by induction on $m$ that
$$
\left.
\begin{array}{lllllll}
(a) \; \lim_{z\rightarrow a}P_{f,a}(z)=\infty_{\mathbb C}\\
(b) \; \lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty\\
\end{array}
\right\}\Longrightarrow \lim_{z\rightarrow a}\Im(P_{f,a}(z))=\infty\;.
$$
Let us suppose now $m=1,\; z=x+iy,\; a=\alpha+i\beta$.
\begin{align*}
P_{f,a}(z)=&\frac{c_{-1}}{z-a}\\
=&\frac{\Re c_{-1}+i\Im c_{-1}}{(z-a)(\overline{z-a})}\overline{z-a}\\
=&\frac{\Re c_{-1}(x-\alpha)+\Im c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2}+
i\frac{\Im c_{-1}(x-\alpha)-\Re c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2}
\end{align*}
Then\begin{align*}
\Re(P_{f,a}(z))=&\frac{\Re c_{-1}(x-\alpha)+\Im c_{-1}(y-\beta)}{(x-\alpha)^2+(y-\beta)^2}\\
=&\frac{\Re c_{-1}}{(x-\alpha)+\frac{(y-\beta)^2}{(x-\alpha)}}+
\frac{\Im c_{-1}}{(y-\beta)+\frac{(x-\alpha)^2}{(y-\beta)}}\;;
\end{align*}
and being by hypotesis $\lim_{z\rightarrow a}\Re(P_{f,a}(z))=\infty$, then at least one between 
$\frac{(x-\alpha)^2}{(y-\beta)}$ and $\frac{(y-\beta)^2}{(x-\alpha)}$ must tend to $0$ as $z\rightarrow a$ (equivalently $(x,y)\rightarrow(\alpha,\beta)$). Then we can conclude, with a simmetry argument, that also $\lim_{z\rightarrow a}\Im(P_{f,a}(z))=\infty$.
Note that the role of $c_{-1}$ is not trivial: we have supposed $\Re c_{-1}\neq0\neq\Im c_{-1}$; the other two cases are easier; if $\Re c_{-1}\neq0=\Im c_{-1}$, suppose wlog $c_{-1}=1$ and $a=0$. Then we're facing the case $P_{f,a}(z)=\frac1{z}$. Now $\frac1{z}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$; by hypotesis $\frac{x}{x^2+y^2}\rightarrow\infty$ as $(x,y)\rightarrow(0,0)$. But $a=0\in\Omega$ hence $\exists t>0$ s.t. $B(0,t[\subseteq\Omega$, so we can argue by simmetry to conclude that also $\frac{y}{x^2+y^2}\rightarrow\infty$ as $(x,y)\rightarrow(0,0)$.
The case $\Re c_{-1}=0\neq\Im c_{-1}$ is similar.
Inductive step: call now
$$
P_{f,a}^{m}=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n}
$$
(we only highlight the $m$). So, let the statement true for $P_{f,a}^{m-1}$ and show it holds for $P_{f,a}^{m}$.
$$
P_{f,a}^{m}
=\sum_{n=1}^{m}\frac{c_{-n}}{(z-a)^n}
=\frac{c_{-m}}{(z-a)^m}+P_{f,a}^{m-1}
$$
so
$$
\Re\left(P_{f,a}^{m}\right)
=\Re\left(\frac{c_{-m}}{(z-a)^m}\right)+\Re\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty
$$
hence exists at least one $k$ in $\{1,\dots,m\}$ s.t. $\Re\left(\frac{c_{-k}}{(z-a)^k}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$:
$\bullet$ if $k\le m-1$ then $\Re\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ and by inductive hyp we have 
$\Im\left(P_{f,a}^{m-1}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ and so also $\Im\left(P_{f,a}^{m}\right)\stackrel{z\rightarrow a}{\longrightarrow}\infty$ which concludes.
$\bullet$ if $k=m$, observing that
$
\frac{c_{-m}}{(z-a)^m}=\frac{c_{-m}}{(z-a)^{m-1}}\frac1{z-a}\;
$
we have
\begin{align*}
\Re\left(\frac{c_{-m}}{(z-a)^m}\right)=&
\Re\left(\frac{c_{-m}}{(z-a)^{m-1}}\frac{1}{z-a}\right)\\
=&\underbrace{\Re\left(\frac{c_{-m}}{(z-a)^{m-1}}\right)}_{\xi_1}
\underbrace{\Re\left(\frac1{(z-a)}\right)}_{\xi_2}-
\underbrace{\Im\left(\frac{c_{-m}}{(z-a)^{m-1}}\right)}_{\eta_1}
\underbrace{\Im\left(\frac1{(z-a)}\right)}_{\eta_2}
\end{align*}
which goes to infinity; hence al least one between $\xi_1,\xi_2,\eta_1,\eta_2$ goes to infinity; if were $\xi_{1}$ and/or $\xi_2$ we fall in the previous case then we conclude.
If otherwise were one of the $\eta$'s then $\Im\left(P_{f,a}^{m}\right)\rightarrow\infty$, so we have finished too.
Then we reached the contradiction , so $a$ can't be a pole.

Case 2: $a$ is an essential singularity.
Simply use Casorati-Weierstra$\beta$ theorem: let $U$ be an open nhb of $a$ in $\Omega$. Then $f(U\setminus\{a\})$ is dense in $\mathbb C$, hence $v$ can't be bounded, absurd.

So we proved that the isolated singularity can't be neither a pole nor an essential singularity. Hence it must be a removable singularity.
A: Update: 
Assume that ${\rm Im}\bigl(f(z)\bigr)<M$ for some $M>0$ and all $z\in\dot\Omega:=\Omega\setminus\{a\}$. The Moebius transform
$$w\mapsto \zeta:={iMw\over 2iM-w}$$
maps the half plane ${\rm Im}(w)<M$ onto the interior of the disc $|\zeta|<M$. It follows that the function
$$g(z):={iM f(z)\over 2iM-f(z)}$$
satisfies
$$|g(z)|<M\qquad\forall z\in\dot\Omega\ .$$
Therefore $g$ has a removable singularity at $a$ and is in fact analytic in all of $\Omega$.  By the maximum principle it follows that $|g(a)|<M$ as well, whence
$$f(z)={2iM g(z)\over iM+g(z)}$$
is analytic in all of $\Omega$.
A: See [Casorati–Weierstrass theorem][1]
If $g(z)$ is bounded for $0<|z-a|<r$ then $g$ has a removable singularity at $z=a$. This is a simple corollary of Cauchy's formula.
