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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...

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up vote 4 down vote accepted

What you were trying to do is fine. The problem is that your function $f(z) = z \bar{z}$ is not complex differentiable, except when $z = 0$. Thus the limit you're trying to compute will only exist when $z = 0$ and that's why you're stuck.

One "easy" way to see that your function is only complex differentiable at $0$ is to use the partial differential operators (aka Wirtinger derivatives)

$$\frac{\partial}{\partial z} = \frac{1}{2} \left ( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right ) \quad \frac{\partial}{\partial \bar{z}} = \frac{1}{2} \left ( \frac{\partial}{\partial x} + i \frac{\partial}{\partial y} \right )$$

Using these operators, a function $f(z)$ is complex differentiable if and only if $\dfrac{\partial f}{\partial \bar{z}} = 0$ (Note that this is just another way of encoding the Cauchy Riemann equations into these operators). In our case, using the fact that the Wirtinger derivatives satisfy the product rule we get

$$\frac{\partial{f}}{\partial \bar{z}} = \frac{\partial{(z\bar{z})}}{\partial \bar{z}} = \frac{\partial z}{\partial \bar{z}} \cdot \bar{z} + z \cdot \frac{\partial{\bar{z}}}{\partial \bar{z}} = 0\cdot \bar{z} + z \cdot 1 = z$$

Thus we see that $\dfrac{\partial{f}}{\partial \bar{z}} = 0 \iff z = 0$, which means that $f$ is complex differentiable only at $z = 0$.

These differential operators are very useful when you're dealing with polynomial functions in $z$ and $\bar{z}$ and you want to see if your function is complex differentiable or not.

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First, we know that the quotient of two differentiable functions is a new differentiable function, assuming the function in the denominator is not 0. So, if $z \bar{z}$ were differentiable at any point other than $z = 0$, since $z$ clearly is differentiable everywhere, we would have that

$$\frac{z \bar{z}}{z} = \bar{z}$$

is differentiable at the same point. But, it is well known that $\bar{z}$ is not differentiable anywhere. (See Example 3.2 here.) Thus, the derivative you are trying to calculate can not possibly exist, except possibly when $z = 0$. In that specific case, the derivative is given by

$$\lim_{z \to 0} \frac{z\bar{z} - 0}{z - 0} = \lim_{z \to 0} \bar{z} = 0$$

since $\bar{z}$ is a continuous function. So, we see that the derivative does exist when $z = 0$. But, it does not exist any where else.

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As Graphth says, it is true that $z\overline z$ is nowhere analytic, so the derivative is not well defined if you try to calculate it as a limit.

However, we can define a differential operator $\frac{d}{dz} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right)$. This operator coincides with the usual complex derivative on the space of holomorphic functions (why?), and it can allow us to make sense of the derivative of functions which would otherwise not be differentiable. In your case, $f(z)=z\overline z = x^2+y^2$, and an easy calculation shows that $\frac{df}{dz} = x-iy=\overline z$.

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Indeed! For that matter, if one imagines that $z$ and $z$-bar are independent variables (in some suitable sense), then it is "obvious" that the derivative is $z$-bar. :) – paul garrett Oct 28 '12 at 16:20
If you mean this, it is best to write $\partial/\partial z$ to alert the reader. – GEdgar Oct 28 '12 at 16:56

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.

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