Question

Find all complex numbers $z$ such that
\[
|z|^2-2\bar z+iz=2i.
\]

Answer 1

the bar is the complex conjugate

Answer 2

|z|^2-2z*+iz = 2i
letting z = x+yi,

x^2+y^2 - 2(x-yi) + i(x+yi) = 2i
x^2+y^2-2x+2yi+ix-y-2i = 0
(x^2+y^2-2x-y)+(x+2y-2)i = 0

so, now just solve

x^2+y^2-2x-y=0
x+2y = 2

x = 2-2y
(2-2y)^2+y^2-2(2-2y)-y=0
5y^2-5y = 0
y = 0 or 1
so, x = 2 or 0

z = 2, i

Answer 3

I am confused on how z = 2 or i

Answer 4

Hi Pumple, it is z = 2, i because z is a complex number so the y is 1 so it is 1i, it is just that the 1 is invisible, but its there