Find all complex numbers $z$ such that
\[
|z|^2-2\bar z+iz=2i.
\]
4 answers
the bar is the complex conjugate
|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
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
I am confused on how z = 2 or i
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
1 answer