\( |z|^2-2\bar z+iz=2i. \)
z*z(bar) - 2(a - bi) + i (a + bi) = 2i
(a + bI) ( a - bi) - 2a + 2bi + ai + bi^2 = 2i
a^2+ b^2 - 2a + 2bi + ai + bi^2 = 2i
a^2 + b^2 - 2a + 2bi + ai - b = 2i
(a^2 + b^2 - 2a - b) + (2b + a)i = 0 + 2i equate coefficients
Thus
a^2 + b^2 - 2a - b = 0 (1)
2b + a = 2 ⇒ a = 2 - 2b ⇒ a = 2 (1 - b) (2)
Sub (2) into (1)
[4 (1 - b)^2] + b^2 - 2 [ 2 (1 - b) ] - b = 0
4 (b^2 - 2b + 1) + b^2 - 4 + 4b - b = 0
4b^2 - 8b + 4 + b^2 - 4 + 4b - b = 0
5b^2 - 5b = 0
b^2 - b = 0
b(b - 1) = 0 set each factor to 0 and solve for b
b = 0 or b = 1
So when b = 0, a = 2 (1 - 0) = 2
And when b = 1, a = 2 ( 1 - 1) = 0
So the solutions are
z = 2 + 0i , z = 0 +1i