Having trouble with proving $\lim_{z\to0}\frac{(Re(z))^2}{|z|} = 0$ using the epsilon definition

46 Views Asked by At

So, I got up to expanding $\frac{(\Re(z))^2}{|z|}$ to $\frac{z^2+2z\overline z + \overline z^2}{4|z|}$ but I don't see other sequences that converge to $0$ to spark inspiration on how to continue.

This one has just really thrown me off. Any tips?

Edit: Actually

$$\frac{(\Re(z))^2}{|z|} = \frac{r^2\cos^2(n)}{|r|e^{in}} = \frac{|r|^2\cos^2(n)}{|r|e^{in}} = \frac{|r|\cos^2(n)}{e^{in}}$$

and $$\lim_{r\to0}\frac{|r|\cos^2(n)}{e^{in}} = 0\frac{\cos^2(n)}{e^{in}} = 0$$

Is it correct?

1

There are 1 best solutions below

0
On

As an alternative we can also use that by $z=x+iy$ and $x\neq 0$

$$\frac{(\Re(z))^2}{|z|}=\frac{x^2}{\sqrt{x^2+y^2}}\le \frac{x^2}{|x|}=|x|\to 0$$