Let zC be such that z^2+3iz-2+i=2+3i.

Question

Let $z\in\mathbb{C}$ be such that $\dfrac{z^2+3i}{z-2+i}=2+3i$. Then the sum of all possible values of $z^2$ is:

Accepted Answer

$$z^2+3i=(z-2+i)(2+3i)=2z-4+2i+3iz-6i-3=(2z-7)+i(3z-4).$$

$z^2-(2+3i)z+(7+7i)=0.$

Treating this as a quadratic in $z$ with roots $z_1,z_2$: sum $=2+3i$, product $=7+7i$.

$$z_1^2+z_2^2=(z_1+z_2)^2-2z_1z_2=(2+3i)^2-2(7+7i)=(4-9+12i)-(14+14i)=-5+12i-14-14i=-19-2i.$$

Solutions · 1