TITLE: How do I find $\frac{\text{d}}{\text{d}z}\left(z\bar{z}\right)$?
QUESTION [8 upvotes]: I am seeking $\frac{\text{d}}{\text{d}z}\left(z\bar{z}\right)$ where $f(z)=z\bar{z}.$
And I know that I need to use the following definition of the derivative:
$$f'(z)=\lim_{\Delta z\to 0}{\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}}.$$
However, I'm not sure if I'm using the definition correctly when I plug in $f(z)$:
\begin{align*}
f'(z)&=\lim_{\Delta z\to 0}{\frac{(z+\Delta z)(\overline{z+\Delta z})-z\bar{z}}{\Delta z}}\\&=\lim_{\Delta z\to 0}{\frac{\overline{\Delta z}(z+\Delta z)+\bar{z}\Delta z}{\Delta z}}\\&=\lim_{\Delta z\to 0}{\frac{\overline{\Delta z}(z+\Delta z)}{\Delta z}}+\lim_{\Delta z\to 0}{\frac{\bar{z}\Delta z}{\Delta z}}\\&=\lim_{\Delta z\to 0}{\frac{\overline{\Delta z}(z+\Delta z)}{\Delta z}}+\bar{z}
\end{align*}
Assuming that I've maneuvered the limit above properly, I'm not sure how to continue from the final line...

REPLY [2 votes]: Your algebraic manipulation is all correct. Now consider the expression
$$
\frac{\overline{\Delta z}}{\Delta z}.
$$
which shows up in your last line. If I take the complex conjugate of this, I get its reciprocal. Therefore it lies on the unit circle (because $w = 1/\overline{w} \Rightarrow ||w||=1$).
As I choose different small complex values of $\Delta z$, this expression is simply the point on the unit circle whose angle with the origin is $-2$ times that of $\Delta z$. In particular, I can get any point on the unit circle I like, no matter how small I make $\Delta z$, so your limit doesn't converge, unless the thing you're multiplying by, namely $z + \Delta z$ tends to 0, i.e. unless $z = 0$, in which case you get 0.