Solve coupled equations by three algebraic methods

From $z-wi=3$, $w=(z-3)/i=3i-zi$. Substitute into the second equation: $z^2-(3i-zi)+6+3i=0$, hence $z^2+iz+6=0$. Its discriminant is $i^2-24=-25$, so $z=(-i\pm5i)/2=2i$ or $-3i$. Then $w=2+3i$ or $-3+3i$, paired respectively.

Write $z=3+wi$. Substitution gives $(3+wi)^2-w+6+3i=0$, or $w^2+(1-6i)w-15-3i=0$. The discriminant is $(1-6i)^2-4(-15-3i)=25$, so $w=(-1+6i\pm5)/2$. This gives $w=2+3i$ with $z=2i$, or $w=-3+3i$ with $z=-3i$.

Again obtain $z^2+iz+6=0$. Let $z=a+bi$, with $a,b\in\mathbb R$. Comparing parts gives $a^2-b^2-b+6=0$ and $a(2b+1)=0$. If $b=-1/2$, then $a^2=1/4-1/2-6=-25/4$, impossible for real $a$. If $a=0$, then $b^2+b-6=(b-2)(b+3)=0$, giving $b=2,-3$. Substitution into $w=-iz+3i$ returns the same two paired solutions.