Find locii of $z$ without putting $z=x+iy$

Question

We have interpret the locii of $z$ for all complex number.

$$\operatorname*{Re}\left(\frac{z+2i}{iz+2}\right) \leq 4 \qquad (z\neq 2i)$$

I know one method by putting $z =x+iy$.

I want to know if there is any other method to do this.

Answer 1

Disclaimer: This doesn't avoid the use of $z = x + iy$, but just makes its application easier.

$$ \Re\left(\frac{z + 2i}{iz + 2}\right) = \Re\left(\frac1i\frac{z + 2i}{z - 2i}\right) = \Im\left(\frac{z + 2i}{z - 2i}\right) $$

Notice that vectors $\vec{u} + \vec{v}$ and $\vec{u} - \vec{v}$ are mutually perpendicular. Hence,

$$ \frac{z+2i}{z-2i} = \frac{|z+2i|}{|z-2i|}(\pm i) $$

which basically means $\frac{z-2i}{z+2i}$ is purely imaginary. So you're down to just finding the locus when

$(1)$ $\frac{z+2i}{z-2i} = \frac{|z+2i|}{|z-2i|}i$, i.e. $z$ lies to the right of the $y$-axis.

$ \frac{|z+2i|}{|z-2i|} \le 4 \\ \implies x² + (y+2)² \le 16x^2 + 16(y-2)² \\ \implies 15x² + 15y² - 68y + 60 \ge 0 $

which is all points on and in the exterior of the right semicircle $15x² + 15y² - 68y + 60 = 0$.

$(2)$ $\frac{z+2i}{z-2i} = -\frac{|z+2i|}{|z-2i|}i$, i.e. $z$ lies to the left of the $y$-axis.

In a similar fashion, you get

$$ 15x² + 15y² - 68y + 60 \le 0 $$

which is all points on and in the interior of the left semicircle $15x² + 15y² - 68y + 60 = 0$.

Answer 2

We have $\Re(\frac {z+2i}{iz +2}) = \Re (\frac {(z+2i)(iz-2)}{(iz+2)(iz-2)}) = \Re (\frac {4z -iz^2 +4i}{z^2 +4}) \leq 4 $. Now, for the condition to be satisfied, we must have $4\Re (z) \leq 4 (z^2 +4) $ $\Rightarrow \Re (z) \leq z\bar z + 4 = |z|^2+4 $ where $\bar z $ is the conjugate of $z $. Hope it helps.