Computing indeterminate complex limits I'm wondering how does one compute limits that are generally computed using L'Hopital's rule for functions of a single real variable. For example, how does one evaluate, the limit,
\begin{align}
\lim_{z \longrightarrow 0} z \log (z), \quad \quad z \in \mathbb{C}
\end{align}
where $\log (z)$ is the principal branch of the logarithmic function. Usually, on the real line, this limit is computed using L'Hopital's rule.
 A: You can do it with real analysis.
\begin{align}
z\cdot \log(z)
&=(a+ib)\cdot\left(\log\left(\sqrt{a^2+b^2}\right)+i\arg(z)\right)\\
&=\left(a\log\left(\sqrt{a^2+b^2}\right)-b\arg(z)\right)+i\left(b\log\left(\sqrt{a^2+b^2}\right)+a\arg(z)\right)
\end{align}
We wish to calculate the limit above as $(a,b)\to(0,0)$. Notice $\arg\big((a,b)\big)$ for the principal branch is the unique $\theta\in(-\pi,\pi)$ with $\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}$.

That said, polar form comes in handy for this particular case. We have
\begin{align}
z\cdot \log(z)
&=re^{i\theta}\cdot\left(\log(r)+i\theta\right)\\
&=r\log(r)\cdot e^{i\theta}+r\cdot i\theta e^{i\theta}
\end{align}
Since $|\theta|\leq \pi$, the term $i\theta e^{i\theta}$ is bounded. We need to calculate the limit of the above expression as $r\to 0$, and this reduces to the real limit of $r\log(r)$ which you can once again deal with using real analysis techniques like L'Hôpital.
A: On the principal branch, the complex logarithm is defined as
$$\log(z)=\log(|z|)+i\text{Arg}(z) \tag1$$
where $-\pi < \text{Arg}(z)\le \pi$.  
Therefore, we can write from $(1)$ that 
$$z\log(z)=z\log(|z|)+iz\text{Arg}(z)\tag2$$
Taking the magnitude of $(2)$ and applying the triangle inequality reveals
$$\begin{align}
\left|z\log(z)\right|&\le |z|\,|\log(|z|)|+|z|\,|\text{Arg}(z)|\\\\
&\le |z|\,|\log(|z|)|+|z|\,\pi \tag 3
\end{align}$$
Noting that $|z|\in \mathbb{R}$, the limit on the right-hand side of $(3)$ can be evaluated using real analysis and in particular, the first term, $|z|\,|\log(|z|)|$ can be evaluated using L'Hospital's Rule. 
