Question

lf the real part of $\frac{z-4}{z-2i}(z\ne 2i)$ is zero then the locus of $z$ is?

• A. $(x-2{)}^2+(y-1{)}^2=25$
• B. $(x+2{)}^2+(y-1{)}^2=5$
• C. $(x-2{)}^2+(y+1{)}^2=5$
• D. $(x-2{)}^2+(y-1{)}^2=5$ excluding the point $(0,2)$

Answer 1

The correct option is D $(x-2{)}^2+(y-1{)}^2=5$ excluding the point $(0,2)$

$Re\left(\frac{z-4}{z-2i}\right)=0$
$\frac{z-4}{z-2i}+\frac{\overline{z}-4}{\overline{z}+2i}=0$
$[\because z+\overline{z}=Re(z)=0]$
$z\overline{z}+2iz-4\overline{z}-8i+z\overline{z}-4z-2i\overline{z}-8i=0$
${\left|z\right|}^2+2i(z-\overline{z})-4(z+\overline{z})=0$
$Letz=x+iy$
$\Rightarrow 2(x^2+y^2)+2i\left(2iy\right)-2\left(2x\right)=0$
$\Rightarrow (x-2{)}^2+(y-1{)}^2=2^2+1^2$
$=5$