## Cauchy-Reimann from first principles

1. The problem statement, all variables and given/known data

Prove from first principles that f(z) = $\overline{z}^2$ is not differentiable at any point z ≠ 0

2. Relevant equations

3. The attempt at a solution

So i guess i Have to show $\stackrel{lim}{h\rightarrow0} \frac{f(z+h)-f(z)}{h}$ is equal to zero right?

$\frac{\overline{z+h}^2-\overline{z}^2}{h}$ but not sure where to go from here.
Do i use z=(x+iy) then $\overline{z} = x-iy$ and so $\overline{z}^2$ = $x^2-y^2-i2xy$
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 Quote by gtfitzpatrick i Have to show $\stackrel{lim}{h\rightarrow0} \frac{f(z+h)-f(z)}{h}$ is equal to zero right?
No, you need to prove that it does not exist. Start by formulating "limit does not exist" in the epsilon-delta language.
 You could assume the limit exists (and hence f is differentiable), and calculate it twice using two sequences which converge to zero, and find that you get two different limits. In this case, your assumption becomes false, so f isn't differentiable. You do know that the definition of differentiability you have is equivalent to having the limit $\lim_{n\rightarrow\infty}\frac{f(z+x_n)-f(z)}{x_n}$ exist for every (complex!) sequence $x_n\rightarrow 0$, right? In this case, I recommend you work with the definition you have, and see what happens when you approach zero along these two sequences: $x_n=1/n$ $y_n=i/n$. In general, such an approach is usually the best way to show that something -isn't- differentiable. Unless of course you have the Cauchy-Riemann equations :)