# Cauchy-Reimann from first principles

gtfitzpatrick

## Homework Statement

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

## 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 $\overline{z}^2$ = $x^2-y^2-i2xy$

voko
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.

christoff
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 :)

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