# Limit of complex quotient with conjugate

This is from the practice GRE online.

Compute $$\lim_{z \to 0} \frac{(\bar{z})^2}{z^2}$$

I let $z = re^{i\theta}$ and computed

$$\frac{(\bar{z})^2}{z^2} = \frac{(re^{-i\theta})^2}{(re^{i\theta})^2} = \frac{1}{e^{4i\theta}}$$ Letting $\theta \to 0$ we see that the limit is 1. But apparently the correct answer is that the limit doesn't exist. Why does this approach fail? Is it because we have to let $r \to 0$ along any $\theta$ path?

• You don't get to let $\theta\to0$. Consider the function $f(z)=z/|z|$ for example. That projects any point onto the unit circle. Geometrically, does it make sense for that to have a limit at $z=0$? No. Sep 17, 2016 at 4:25

You should be letting $r\to 0$, not $\theta\to 0$. So you have shown that the result depends on the direction $\theta$ of approach, so the limit does not exist.
You're right that along the line $\theta=0$ (the positive side of the $x$-axis) $f(z)=(\overline{z})^2/z^2$ is identically 1. But along the line $\theta=\pi/4$ we have $f(z)=1/e^{4i\theta}=-1$. In order for the limit to exist, $f$ would have to approach the same value no matter how $z$ approaches the origin. Since $f$ tends to 1 along your path to the origin and tends to $-1$ along mine, the limit must not exist.