Show that $\frac{z-1}{\mathrm{Log(z)}}$ is holomorphic off $(-\infty,0]$ Let $f(z)=\frac{z-1}{Log(z)}$ for $z\neq 1$ and $f(1)=1$. Show that $f$ is holomorphic on $\mathbb{C}\setminus(-\infty,0]$. 
I know it looks like an easy problem, but I got stuck and need some clarification. The way I see it, I need to show that $f$ is complex differentiable at every point in $\Omega=\mathbb{C}\setminus(-\infty,0]$. So, if we take any $z_{0}$ different than $1$, our function is the quotient of two complex differentiable functions on $\Omega$, so is complex differentiable ($Log(z_{0})\neq0)$. Now, if we take $z_{0}$ to be $1$, than if $f$ is continous at that point, we could use the Cauchy-Riemann equations to check whether $u_{x}(1,0)=v_{x}(1,0)$ and $u_{y}(1,0)=-v_{x}(1,0)$ with $f=u+iv$. My question is: Is this the fastest way to show complex differentiability at $1$? I mean, how do I get to differentiating $u$ and $v$ if I don't know them explicitly? I would also appreciate it if someone could give me some hints on how to compute the limit of $f$ at $1$. 

If $g$ is holomorphic on an open set $G$, $a\in G$, and $g(a)=0$, then there is a positive integer $n$ and a holomorphic function $h$ on $G$ such that $g(z)=(z-a)^nh(z)$ for all $z\in G$, and $h(a)\neq 0$.  (This $n$ is the multiplicity of the zero of $g$ at $a$.)
Since $\mathrm{Log}(1)=0$, and $\mathrm{Log}'(1)=1\neq 0$, there exists a holomorphic function $h$ on $\mathbb C\setminus(-\infty,0]$ such that $\mathrm{Log}(z)=(z-1)h(z)$ for all $z$, and $h(1)\neq 0$.   Since $\mathrm{Log}$ has no zeros except at $1$, $h$ has no zeros.  Therefore $f=\dfrac{1}{h}$ is holomorphic on $\mathbb C\setminus(-\infty,0]$.
