Simplification of complex conjugates

Question

I was working through a limit problem and got stuck on the simplification. I have the answer key however I am not sure where I can simplify more.

So the question is, $$f(z) = z \bar z$$

Differentiable at z=0.

I know I can take the limit of this as it approaches zero to check for this.

$$ f'(z) = \frac{f(z+h) - f(z)}{h}$$

Then the equation becomes:

$$ f'(z) \lim z \Rightarrow 0 = \frac{(h+z)^2 (\bar h +\bar z) - z^2\bar z}{h}$$ $$ f'(z) \lim z \Rightarrow 0 = \frac{ (z^2 \bar h) + 2zh\bar h + h^2 \bar h + 2z \bar z h + h^2 \bar z}{h}$$

The professor has simplified then solution to: $$f'(z) \lim z \Rightarrow 0 = \frac{z^2 \bar h + 2z \bar z h }{h} $$

I think I must be missing some sort of an identity in my base knowledge. Any guidance would be greatly appreciated.

Answer 1

We have that

$$(h+z)^2 (\bar h +\bar z) - z^2\bar z=(h^2+2hz+z^2)(\bar h +\bar z) - z^2\bar z=$$

$$=h^2\bar h+2h\bar hz+z^2\bar h+h^2\bar z+2hz\bar z+z^2\bar z- z^2\bar z=$$

$$=h^2\bar h+2h\bar hz+h^2\bar z+z^2\bar h+2z\bar zh=z^2\bar h+2z\bar zh \iff h^2\bar h+2h\bar hz+h^2\bar z=0$$

and as noticed by Lord Shark it seems we are neglecting the higher order terms for $h$ (and $\bar h$) since we are assuming $|h|<<|z|$.

Answer 2

I have got $$\overline{h}h^2+\overline{h}z^2+2h\overline{h}z+h^2\overline{z}+2hz\overline{z}$$