Question

If the imaginary part of $\frac{z-4}{z-2i}$ is $0$ , then the locus of $z$ is

• A. Ellipse
• B. Circle
• C. Straight line
• D. Parabola

Answer 1

The correct option is C Straight line

To find locus of $z$
If the imaginary part of $\frac{z-4}{z-2i}$ is $0$
We have,
$\frac{x+iy-4}{x+iy-2i}$
$\frac{x-4+iy}{x+i(y-2)}$

$=\frac{\left(\right(x-4)+iy)(x-(y-2\left)i\right)}{[x+i(y-2\left)\right][x-i(y-2\left)\right]}$
$=\frac{x^2+y^2-4x-2y+i(2x+4y-8)}{x^2+(y-2{)}^2}$

$=\frac{x^2+y^2-4x-2y}{x^2+(y-2{)}^2}+i\frac{(2x+4y-8)}{x^2+(y-2{)}^2}$
If the imaginary part is $0$, then we have

$\frac{(2x+4y-8)}{x^2+(y-2{)}^2}=0$

$2x+4y-8=0$
which represents equation of a straight line.
Hence , Option C