Question

Determine the locus of $z,z\ne 2i$ such that $Re\left(\frac{z-4}{z-2i}\right)=0$.

Answer 1

Let $z=x+iy$.

Now,

$\left(\frac{z-4}{z-2i}\right)$

$=\frac{(x-4)+iy}{x+i(y-2)}$

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

$=\frac{x(x-4)+y(y-2)+i\{xy-(x-4\left)\right(y-2\left)\right\}}{x^2+(y-2{)}^2}$

$=\frac{x(x-4)+y(y-2)+i\{4y+2x-8\}}{x^2+(y-2{)}^2}$......(1).

Given $Re$ $\left(\frac{z-4}{z-2i}\right)$$=0$.

Then from (1) we've,

$4y+2x-8=0$

or, $x+2y=4$.

This represents the equation of al line.