Q:

Show that $f(z):=\frac{e^z}{1-z}$ is a holomorphic function on $\mathbb C\setminus${$1$}

If we have $f:\mathbb C\setminus${$1$}$\to\mathbb C$ given by 
$f(z):=\frac{e^z}{1-z}$,
how can we show that $f$ is a holomorphic function on $\mathbb C\setminus${$1$}?
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I understand that for a function to be holomorphic on a domain, it means that it is complex differentiable in this domain, but how do I write this mathematically?
Also, how do I incorporate the $\mathbb C\setminus${$1$} aspect into the answer, as logically it is clearly seen that $z$ cannot take the value of $1$, or else the denominator of $f$ would be zero, but how would I formally add this to my proof? Thanks!

A:

I suggest you prove (or look up the proof of) a more general fact:
$``$If $g:U \to \mathbb{C}$ and $h:U \to \mathbb{C}$ are holomorphic, then $z\mapsto g(z)/h(z)$ is holomorphic on the set
$$U\setminus\{w \in U \, : \, h(w)=0\}."$$
In your case, we have $g(z)=e^z$ and $h(z)=1-z.$
It is well-known that $g$ is holomorphic everywhere on $\mathbb{C}.$
It is easily seen that $h$ is also holomorphic everywhere on $\mathbb{C}.$
The only point $w \in \mathbb{C}$ with $h(w)=0$ is $w=1.$
Therefore, by the above result, $z\mapsto e^z/(1-z)$ is holomorphic everywhere on $\mathbb{C}\setminus\{1\}.$

