How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$? How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$?
I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$
And that I could not get out, can anyone help me?
 A: Since the modulus of $\dfrac{\bar z^2}{z}$ is $|z|$, the limit is $0$.
A: From your calculation :
$$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$
$$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$
From here, show that both the limits are zero by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$.
For the first limit,
$$\left|\frac{r^3\cos^3\theta-3r^3\cos \theta\sin^2\theta}{r^2}\right|\le 4r<\epsilon$$whenever, $r^2<\epsilon^2/16$ i,e, whenever $|x|<\epsilon/\sqrt 8=\delta(say)$ , and $|y|<\epsilon/\sqrt 8=\delta(say)$.
Similarly the second limit will be zero and hence the given limit will be $0$.
A: Let $z=re^{i\theta}$. Then this equals:
$$\lim_{r\to 0} \frac{(re^{-i\theta})^2}{re^{i\theta}}=\lim_{r\to 0} \frac{r^2e^{-2i\theta}}{re^{i\theta}}=\lim_{r\to 0} r e^{-3i\theta}=0$$
A: How about
$$
\lim_{z\to 0} \frac{\bar{z}^2}{z} =
\lim_{z\to 0} \frac{\bar{z}^2z^2}{z^3} =
\lim_{z\to 0} \frac{\lvert z\rvert^4}{z^3} = 
\lim_{z\to 0} \frac{\lvert z\rvert^4}{\lvert z \rvert^3 e^{3i\phi(z)}}
= 0
$$
