Describe the domain in the plane $\mathbb{R}^2$

Question

Please check if I do it correctly:

Let $D=\{z\in\mathbb{C}: \left|\frac{2zi-1}{z+2i}\right|<1\}$. Describe $D$ in the complex plane.

Naively, I just try to solve the equation, using the fact that it is not well-defined at $z=-2i$:

$\left|\frac{2zi-1}{z+2i}\right|<1$ iff $\left|2zi-1\right|<\left|z+2i\right|$ iff $\left|2zi-1\right|^2<\left|z+2i\right|^2 $ iff $ \left|2zi\right|^2-2Re(2zi)+1<\left|z\right|^2-2Re(2zi)+\left|2i\right|^2 $ iff $\left|z\right|^2 <1$. So, $D$ is just the open unit ball.

Is it okay?

Answer 1

Yes, you are correct. More generally, if $|a|>1$ then $$\left|\frac{\bar{a}{z}-1}{z-a}\right|<1\Leftrightarrow |z|<1,$$ because $$\begin{align} \left|\frac{\bar{a}{z}-1}{z-a}\right|<1 &\Leftrightarrow |\bar{a}{z}-1|^2<|z-a|^2\Leftrightarrow |a|^2|z|^2-2\text{Re}(\bar{a}{z})+1<|z|^2-2\text{Re}(\bar{a}{z})+|a|^2\\&\Leftrightarrow (1-|a|^2)(1-|z|^2)<0 \Leftrightarrow |z|<1.\end{align}$$ In a similar way, if $|a|<1$ then $$\left|\frac{\bar{a}{z}-1}{z-a}\right|<1\Leftrightarrow |z|>1.$$

Answer 2

As pointed out it is correct, note that for the final step we have

$$\left|2zi-1\right|^2<\left|z+2i\right|^2 \iff \left(2zi-1\right)\,\left(-2\bar zi-1\right)<\left(z+2i\right)\,\left(\bar z-2i\right)$$

from which the result follow.

Answer 3

It is O.K.

My approach: let $T(z)=\frac{2zi-1}{z+2i}$, $T$ is a Moebius transformation. Let $\mathbb D$ denote the open unit disc.

It is easy to see that $|T(1)|=|T(-1)|=|T(i)|=1$ and $|T(0)|<1$.

Hence: $T(\mathbb D)=\mathbb D$.