Question

Given that the complex number z and its conjugate z* satisfy the equation zz* + 2iz = 12 +6i. Find the possible value of z.

Answer 1

$z=x+iy$

$z^*=x-iy$

$zz^*=x^2+y^2$

$x^2+y^2+2xi-2y=12+6i$

$x^2+y^2-2y=12$

$2xi=6i$

$x=3$

$9+y^2-2y=12$

$y^2-2y-3=0$

$y_1=\frac{2-\sqrt{4+12}}{2}=-1$

$y_2=3$

Answer:

$z_1=3-i$

$z_2=3+3i$