If $ f $ has pole at $0$ then show that $e^f$ can't have pole at $0$. i am trying to show that if  $ f $ has a pole at $0$ then   $ e^f $ can't have removable singularity at $0$ ? I tried to show that but i have a problem .
I assume that $e^f$ has removable singularity . 
Therefor $\lim {e^{f(z)}}=c $ while $z->0$, where c is a complex number .              If $ c\ne 0$ then we can say that $(e^{f(z)})'/e^{f(z)} $ has removable singularity because the $ \lim (e^{f(z)})'/e^{f(z)}$ exists but $(e^{f(z)})'/e^{f(z)} =(f(z))' $ which it leads us to a contradiction,because if $f$ has pole then $ f ' $ has pole too. 
But if $c=0$ i can't see where it fails, because $(e^{f(z)})'/e^{f(z)} $ might have  pole at $0$.
Thank you in advance! 
 A: Put $g(z)=\exp(f(z))$, and suppose that $g$ has a removable singularity at $0$. Then you can write $g(z)=z^m h(z)$, with $m$ integer $m\geq 0$ ($m=0$ is the case you have done) and $h$ analytic at $0$,  $h(0)\not =0$. Then $$ \frac{g^{\prime}(z)}{g(z)}=f^{\prime}(z)=\frac{m}{z}+\frac{h^{\prime}(z)}{h(z)}$$ show that if $m=0$, $f^{\prime}$ has no pole at zero, a contradiction; and if $m\geq 1$, that $f^{\prime}$ has a pole of order $1$ at $0$. But as $f$ has a pole at $0$, say of order $s$, $f^{\prime}$ has a pole of order $s+1\geq 2$ at zero, contradiction again. (In fact you can take $m$ in $\mathbb{Z}$, and show in this way that $\exp(f(z))$ cannot have a pole at $z=0$) 
A: [EDITED]
Suppose $f$ has a pole at $0$.  For any $\epsilon > 0$, let $U_\epsilon = \{z: 0 < |z| < \epsilon\}$.  By the Open Mapping Theorem, $(1/f)(U_\epsilon) = \left\{\dfrac{1}{f(z)}: z \in U_\epsilon\right\}$ contains a deleted neighbourhood of $0$, so $f(U_\epsilon)$ contains a deleted neighbourhood of $\infty$, i.e. 
$\{w: |w|>R\}$ for some $R > 0$.  In particular, it contains the strip $S_R = \{w: R < \text{Im}(w) \le R + 2 \pi\}$.  Since $\exp(S_R) = \mathbb C \backslash \{0\}$, we see that $e^f$ takes all nonzero complex values on a deleted neighbourhoood of $0$.     Thus the singularity of $e^f$ at $0$ must be essential.
