Limit of complex number over its conjugate as it approaches zero

Question

Basically, I have to look at this thing: $$\lim_{z\to 0}{\frac{z}{z^*}}=1$$ I know this, but I need to prove it. I tried using the following approach. $$\frac{re^{i\theta}}{re^{-i\theta}}=e^{2i\theta}$$ Unless I can get a definite value for $\arg 0$, I'm stuck at an impasse, as this makes it clearly look like the limit doesn't even exist since the angle of attack for this limit changes its value. What am I missing? I have to be missing something here in this hot mess.

Answer 1

You are correct. The fraction is dependent to $\theta$ and the limit does not exist and is not equal to $1$.

Answer 2

The way you are proceeding is completely correct. Just add the conclusion.

We have $$\lim_{z\to 0}{\frac{z}{z^*}}$$

Now Z is in the complex plane. Let's move toward zero through the imaginary axis.

as on imaginary axis, z is imaginary always, so $z=-z^*$

Hence $$\lim_{z\to 0}{\frac{z}{z^*}}=\frac{z}{-z}=-1$$

Now let's approach zero from the real axis, on the real axis we have

$z=z^*$

Hence $$\lim_{z\to 0}{\frac{z}{z^*}}=\frac{z}{z}=1$$

Conclusion: If we approach zero along different paths then we get different limit hence the limit at zero doesn't exist.

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In your case: The way you proceeded,

We know that that $z\to 0$ is to say $r\to 0$, and no condition on $\theta$

$$\begin{align} \frac{z}{z^*} = e^{-2i\theta} \end{align}$$

But here limit depends on $\theta$, that is depends on how z approaches zero. In another word, if we choose the different direction of approach we'll get different values of limit.

Note: the different value of limits that we are getting will always lie on a circle of unit radius around the center.