Why $a$ is not an essential singularity and how $f$ can be extended in our online lecture we had the following statement but I don't see exactly why it's true , if anyone could explain me why  it would be appreciated.


We have  $a \in \mathbb{C}$ a complex number and $r \in \mathbb{R}$ with $r>0$. Let $U=\mathbb{D}(a,r)\ _{\diagdown \left\{a \right\}}$ and $f \in O(U)$ a holomorphic function such that $Re(f(z)) \geq 0 \; \forall z \in U$.


And it says that $f$ can be extended ( in a holomorphic function )  all over the disk $\mathbb{D}(a,r)$ and that $a$ isn't a essential singular point. 
Thanks in advance for your help .

The singularity of $f$ at $a$ is either a pole, a removable singularity or an essential singularity.  
If it were a pole of order $n$, you'd have $f(z) = c (z-a)^n + O((z-a)^{n-1})$ for some $c \ne 0$.  By having $z$ approach $a$ from a direction such that $c (z-a)^n$ is on the negative real axis you'd get $\text{Re}(f(z)) < 0$.
If it were an essential singularity, Casorati-Weierstrass theorem would say there is $z$ near $a$ such that $f(z)$ is near, say, $-1$.
The only other possibility is a removable singularity.  That says that $f$ can be extended to be analytic at $a$.
