does the limit exist for the following:
$$\lim_{z \to 0} \frac {Re(z^2)}{|z|^2}$$
My take: I tried to substitute $z=x+iy$ for z and then solve the limit but I get 0. According to the wolfram alpha its limit is 1; so can someone please explain this to me? What am I doing wrong here?
Thank You!!!
Best Answer
If $z \in \mathbb R$ and $z \ne 0$, then $\frac {Re(z^2)}{|z|^2}=1$
and
if $z \in i\mathbb R$ and $z \ne 0$, then $\frac {Re(z^2)}{|z|^2}=-1$
Conclusion ?