For$\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$
- Does $\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$ exist? If the answer is no, why?
- Does $\bar{z}$ represents $a-bi$?
For$\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}$
If $z = a + bi$, then $\bar{z} = a - bi$. Equivalently, if $z = r e^{i\theta}$, $\bar{z} = r e^{-i \theta}$.
Take $z = re^{i\theta}$ and as $z \to 0$, we have $r \to 0$.
Plug in $z = r^{i\theta}$ in $\dfrac{\bar{z}^2}{z^2}$. Now, let $r \to 0$ and see what happens for different $\theta$'s.