Question

If $z=x+iy,w=\frac{2-iz}{2z-i}$ and $|w|=1,$ find the locus of z and illustrate it in the complex plane.

Answer 1

${\left|\omega \right|}^2=\omega \overline{\omega }=\left(\frac{2-iz}{2z-i}\right)\left(\frac{2-\overline{i}\overline{z}}{2\overline{z}-\overline{i}}\right)=\left(\frac{2+i\overline{z}}{2\overline{z}+i}\right)\left(\frac{2-iz}{2z-i}\right)$

$1=\left(\frac{4-2iz+2i\overline{z}-i^{2}z\overline{z}}{4z\overline{z}-2i\overline{z}+2iz-i^2}\right)=\frac{4-2i(x+iy)+2(x-iy)-i^2{\left|z\right|}^2}{4{\left(|z|\right)}^2-2i(x-iy)+2i(x+iy)+1}$

$4(x^2+y^2)-2ix+2i^{2}y+2ix+2i^{2}y+1=4-2ix-2i^{2}y+2ix-2i^{2}y+1(x^2+y^2)$

$\to 3(x^2+y^2)-3+4i^{2}y=-4i^{2}y$

$\to 3(x^2+y^2)-3+8i^{2}y=0\Rightarrow 3(x^2+y^2)-3-8y=0$

$\to x^2+y^2-\frac{8y}{3}-1=0$

$\to x^2+{(y-\frac{4}{3})}^2=1+\frac{16}{9}=\frac{25}{9}={\left(\frac{5}{2}\right)}^2$

$\to |z-\frac{4i}{3}|=\frac{5}{2}$