Question

If Z is a complex number such that
Z - 2i I = I Z + 2i I , show that Im(Z) = 0.

Answer 1

$let \: z \: = x + iy. \: \: then \\ |z - 2i| = |z + 2i| \: becomes \\ |x + (y - 2)i| = |x + (y + 2)i| \\ {x}^{2} + {(y - 2)}^{2} = {x}^{2} + {(y + 2)}^{2} \\ simplifying \: \: y = 0 \\ i.e. \: \: im(z) = 0$

Answer 2

We know that complex numbers are in the form of

 $z=x+iy$

Given

$|z-2i|=|z+2i|$

Now we will substitute the value of $z$ in the given equation.

$|x+iy-2i|=|x+iy+2i|\\|x+i(y-2)|=|x+i(y+2)|\\x^{2} +(y-2)^{2}=x^{2} + (y+2)^{2}\\(y-2)^{2}=(y+2)^{2}\\y^{2} +4-4y=y^{2}+4+4y\\-4y=4y\\8y=0\\y=0\\ie \\Img(z)=0$

Hence we proved that

$Img (z)=0$