Simplify $\lim_{z\to1}$ $\frac{\text{Log}(z^2)}{z-1}$ $z$ is a complex number $x+iy$
$\text{Log}(z)=\ln(|z|)+i\theta$, where $\theta$ is the angle that $z$ makes off the real axis and $-\pi<\theta\leq\pi$
I was trying to help someone with this problem, but I didn't really see how it could really be simplified or solved for.  I feel like I'm missing something obvious.  Any help would be appreciated. 
 A: METHODOLOGY $1$:  Use the definition of the derivative
Note that the complex logarithm as defined on the principal branch is analytic in a neighborhood of $z=1$.  
Inasmuch at $\log(1)=0$ on the principal branch, the limit $\lim_{z\to 1}\frac{\log(z^2)}{z-1}=\lim_{z\to 1}\frac{\log(z^2)-\log(1)^2}{z-1}$ is simply the derivative of $\log(z^2)$ at $z=1$.  Proceeding, we have
$$\begin{align}\lim_{z\to 1}\frac{\log(z^2)}{z-1}&=\lim_{z\to 1}\frac{\log(z^2)-\log(1)^2}{z-1}\\\\
&=\left.\left(\frac{d}{dz}\log(z^2)\right)\right|_{z=1}\\\\
&=\left.\left(\frac2z\right)\right|_{z=1}\\\\
&=2
\end{align}$$ 
as expected.

METHODOLOGY $2$:  Use the series representation of the complex logarithm
For $|z-1|\le 1$, and $z\ne 0$, $\log(z)=\sum_{n=1}^\infty \frac{(-1)^{n-1}(z-1)^n}{n}$.  Hence, we can write for $|z^2-1|\le 1$, $z\ne0$
$$\log(z^2)=\sum_{n=1}^\infty \frac{(-1)^{n-1}(z^2-1)^n}{n}=\sum_{n=1}^\infty \frac{(-1)^{n-1}(z-1)^n(z+1)^n}{n}$$
Dividing by $z-1$, and letting $z\to 1$ from inside the unit circle, we have
$$\lim_{z\to 1}\frac{\log(z^2)}{z-1}=2\tag1$$
We can continue the logarithm analytically outside the unit circle in a neighborhood of $z=1$ and arrive at $(1)$.
A: We can use L'Hopitals rule for complex differentiable functions. If $f$ and $g$ are two functions that are complex differentiable at $c\in\mathbb{C}$, then
$$\lim_{z\to c}\frac{f\left(z\right)}{z-c}=f^{\prime}\left(c\right)\text{ and }\lim_{z\to c}\frac{g\left(z\right)}{z-c}=g^{\prime}\left(c\right),$$
which gives us that
$$\lim_{z\to c}\frac{f\left(z\right)}{g\left(z\right)}=\lim_{z\to c}\frac{\frac{f\left(z\right)}{z-c}}{\frac{g\left(z\right)}{z-c}}=\frac{f^{\prime}\left(c\right)}{g^{\prime}\left(c\right)}.$$
Assuming that you already know that $\operatorname{Log}\left(z^{2}\right)$ and $z-1$ are both complex differentiable at $z=1$ (it may be easier to show that $\operatorname{Log}\left(z^{2}\right)$ is complex differentiable using the Cauchy-Riemann equations in polar form or that it is the composition of two holomorphic functions), then
$$\lim_{z\to1}\frac{\operatorname{Log}\left(z^{2}\right)}{z-1}=\lim_{z\to1}\frac{2z}{z^{2}}=\lim_{z\to1}\frac{2}{z}=2.$$
