+612 Find all complex numbers $z$ such that \[ |z|^2-2\bar z+iz=2i. \]
+129852 \( |z|^2-2\bar z+iz=2i. \)
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
+26393 Find all complex numbers $z$ such that
\(|z|^2-2\bar z+iz=2i.\)
|z|^2-2\bar z+iz=2i.
\(\begin{array}{|rcll|} \hline |z|^2-2\bar z+iz &=& 2i \quad & | \quad |z|^2=z\bar z \\ z\bar z-2\bar z+iz &=& 2i \quad & | \quad -iz \\ z\bar z-2\bar z &=& 2i -iz \\ \bar z(z-2) &=& i(2 - z) \\ \bar z(z-2) &=& -i(z - 2) \quad & | \quad z-2 \ne 0 \\ \bar z &=& -i\left(\dfrac{z - 2}{z-2}\right) \\ \bar z &=& -i \quad & | \quad \bar z = a-bi \\\\ a-bi &=& 0-i \quad & | \quad \text{compare both sides} \\ && \boxed{a=0\qquad b = 1 \qquad \text{or} \qquad z= 0 + 1i} \\\\ \bar z(z-2) &=& -i(z - 2) \quad & | \quad z-2 = 0 \\ z-2 &=& 0 \quad & | \quad + 2 \\ z &=& 2 \quad & | \quad z = a+bi \\\\ a+bi &=& 2+0i \quad & | \quad \text{compare both sides} \\ && \boxed{a=2\qquad b = 0 \qquad \text{or} \qquad z= 2 + 0i} \\ \hline \end{array}\)
