Question

Describe all solutions to$$zw -3w - 2iw + 4iz = - 8 + 12i$$where $z$ and $w$ are complex numbers (i is the imaginary number)

Answer 1

Rewrite the equation as
w(z - 3 - 2i) + 4i z = -8 + 12i.
Bring the constant term to the left:
w(z - 3 - 2i) = -4i z - 8 + 12i = -4i(z - 3 - 2i),
since -4i(z - 3 - 2i) = -4iz -8 +12i.

If z ≠ 3 + 2i, divide by z - 3 - 2i to get w = -4i.

If z = 3 + 2i, the left factor vanishes and the equation reduces to 4i(3+2i) = -8 + 12i, which holds, so w can be any complex number.

Hence the solution set is
{(z,w): z ≠ 3+2i, w = -4i} ∪ {(3+2i, w): w ∈ C}.