how to prove that a function is not complex differentiable

Question

I was working on a problem on the complex differentiability of the following function:

$f(z)= z \operatorname{Re}(z)$. How to find the points where the given function is not differentiable. My effort:

$f(z)= z\operatorname{Re}(z) = zx$ where $x$ is the real part of $z$. No after applying the differentition formula i got \begin{eqnarray*} \frac{f(z+\Delta z) - f(z)}{\Delta z} &=&\frac{(z+\Delta z)Re(z+\Delta z)-zx}{\Delta z} &=&\frac{(z+\Delta z)(x+\Delta x)-zx}{\Delta z} \end{eqnarray*}

with the limit $\Delta z$ tends to $0$. Now how do I find the points where function is not differentiable? Kindly help. Any suggestion or hint will be helpful.

Answer 1

Just continue on from where you stopped:

$$ \begin{eqnarray*} \frac{f(z+\Delta z) - f(z)}{\Delta z} &=&\frac{(z+\Delta z)Re(z+\Delta z)-zx}{\Delta z}\\ &=&\frac{(z+\Delta z)(x+\Delta x)-zx}{\Delta z}\\ &=&x+\Delta x+\frac{z\Delta x}{\Delta z}\\ \end{eqnarray*} $$

If $z\ne 0$, then limit $\Delta z\to 0$ along the real line gives $x+z$ but the same along with imaginary line gives $x$, and they are not equal. If $z=0$, then it is obvious that the limit is $x$.