Simplify $\lim_{z\to1}$ $\frac{\text{Log}(z^2)}{z-1}$

Question

$z$ 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.

Answer 1

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.

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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 $(1)$.

Answer 2

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 $$\lim_{z\to1}\frac{\operatorname{Log}\left(z^{2}\right)}{z-1}=\lim_{z\to1}\frac{2z}{z^{2}}=\lim_{z\to1}\frac{2}{z}=2.$$