Finding limit of a complex function

Question

I am asked to show that limiting value of $\dfrac{\bar z^2}z$ is $0$ using $\varepsilon$-$\delta$ definition when z tends to 0.

Is it alright to take $\delta=\varepsilon$ and $\varepsilon>|z|$?

As we get $$\frac{\left|z\right|^{\,2}}{|z|}<\varepsilon\implies\frac{\left|\bar z\right|^2}{|z|}<\varepsilon\implies\left|\frac{\bar z^2} z\right|<\varepsilon$$ whenever $|z|<\delta$. How can we clearly say that |z| is always greater than 0 as epsilon needs to be greater than 0.

Answer 1

See this

$$\Big| \frac{{\bar z}^2} {z}\Big| = \frac{|{\bar z}^2|} {|z|} = \frac{|{z}|^2} {|z|} =|z|<\epsilon =\delta. $$

So we have

$$|z|<\delta \implies \Big| \frac{{\bar z}^2} {z}\Big|< \epsilon. $$

Note:

$$|\bar{z} |^2 =|z|^2. $$

Answer 2

We use the following relation: $$\vert z^2\vert =z\bar{z} \implies \vert\bar{z}^2\vert=\bar{z}z$$

So we see that $$\vert\frac{\bar{z}^2}{z}-0\vert=\vert \bar{z}\vert$$ and for $z=x+iy$ going to $0$ we also have $\bar{z}=x-iy$ going to zero, so we get:

$$\lim_{z \to 0} \frac{\bar{z}^2}{z}=0$$

To see this using $\epsilon - \delta$ definition:

Let $\epsilon >0$ arbitrary. Then we can pick $\delta = \epsilon$ and we need to check that $ \forall z \in \mathbb C $ such that $\vert z \vert < \delta$ we get $\vert \frac{\bar{z}^2}{z}-0\vert< \epsilon$. Now by calculations done above we see that this is indeed the case.