Solve coupled equations involving a conjugate

The ratio requires $z\ne0$ and gives $w=(3/2)iz$. Substitution yields $z^*-3iz=4+4i$. Let $z=x+iy$; then $x-iy-3i(x+iy)=x+3y+(-y-3x)i$.

Compare parts: $x+3y=4$ and $-y-3x=4$. Multiply the first by 3 to get $3x+9y=12$; adding the second gives $8y=16$, hence $y=2$. Substitute into $x+3y=4$: $x=-2$. Thus $z=-2+2i$.

$w=(3/2)i(-2+2i)=-3-3i$. Check $w/z=(3/2)i$ and $z^*-2w=(-2-2i)-2(-3-3i)=4+4i$. The source’s next line after $y=2$ is a substitution giving $x=-2$; a duplicated “8y=16” in the extraction is a text-extraction artifact, not an extra printed equation.