Solve for y
y=-iz
Solve for z
z=iy
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z=\frac{\left(2+3i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}y
Multiply both numerator and denominator of \frac{2+3i}{3-2i} by the complex conjugate of the denominator, 3+2i.
z=\frac{13i}{13}y
Do the multiplications in \frac{\left(2+3i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}.
z=iy
Divide 13i by 13 to get i.
iy=z
Swap sides so that all variable terms are on the left hand side.
\frac{iy}{i}=\frac{z}{i}
Divide both sides by i.
y=\frac{z}{i}
Dividing by i undoes the multiplication by i.
y=-iz
Divide z by i.
z=\frac{\left(2+3i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}y
Multiply both numerator and denominator of \frac{2+3i}{3-2i} by the complex conjugate of the denominator, 3+2i.
z=\frac{13i}{13}y
Do the multiplications in \frac{\left(2+3i\right)\left(3+2i\right)}{\left(3-2i\right)\left(3+2i\right)}.
z=iy
Divide 13i by 13 to get i.
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