Question

Find the points of the plane which satisfy the following equations.
$\left|\frac{z+i}{z-3i}\right|=1.$

Answer 1

$\left|\frac{z+i}{z-3i}\right|=1$

a- point of plane which satisfy

$|z+i|=|z-3i|$

Suppose $z=x+iy$

Here (x, y) are points of the plane which satisfy the equation.

$|x+iy+i|=|x+iy-3i|$

$\Rightarrow |x+i(y+1)|=|x+i(y-3)|$

$(\because |z|=\sqrt{z\overline{z}}$ if $z=a+ib$ then $\overline{z}=a-ib)$

so that

$\sqrt{\{x+i(y+1\left)\right\}\{x-i(y+1\left)\right\}}=\sqrt{\{x+i(y-3\left)\right\}\{x-i(y-3\left)\right\}}$

$\Rightarrow \sqrt{x^2+(y+1{)}^2}=\sqrt{x^2+(y-3{)}^2}$

$\Rightarrow x^2+(y+1{)}^2=x^2+(y-3{)}^2$

$\Rightarrow y^2+2y+1=y^2+9-6y$

$\Rightarrow 8y=8$

$\Rightarrow y=1$

So $y=1$ & $x\in R$

Here points of the plane which satisfy the equation are all point on the line $y=1$.