Question 768565
{{{zi=-4+2-3i}}}
{{{zi=-2-3i}}}
{{{z=-2/i-3i/i}}}
{{{z=(-2-3i)/i}}}
{{{z=((-2-3i)/(i))((-i)/(-i))}}}
{{{z=(-2i+3i^2)/(-1*(-1))}}}
{{{z=-2i-3}}}
{{{highlight(z=-3-2i)}}}