# Complex numbers problem!

• January 2nd 2010, 06:59 AM
Complex numbers problem!
How to find complex numbers in form z=x+yi which satisfies :

• January 2nd 2010, 07:35 AM
Defunkt
Let $z=x+iy$ be a solution of this equation, then:

$(x+iy)(x-iy) -2i = 2i(x-iy) \Rightarrow (x-iy)(x+iy-2i)=2i$

Now,

$(x-iy)(x+i(y-2)) = x^2-2xi+y^2-2$

Substitute back and get: $x^2+y^2-2-2xi = 2i \rightarrow -2x = 2 \Rightarrow x=-1$

$x^2+y^2=2\Rightarrow y^2=1 \Rightarrow y=\pm 1$

So our solutions are:
$z_1=-1+i$
$z_2=\bar{z_1} = -1-i$
• January 2nd 2010, 11:49 AM