Question

If complex number $\frac{1+2i+3i^2}{1-2i+3i^2}$ can be written in the form of $a+ib,$ then $a\cdot b=$

Answer 1

Let $z=\frac{1+2i+3i^2}{1-2i+3i^2}$
$\Rightarrow z=\frac{1+2i-3}{1-2i-3}$
$\Rightarrow z=\frac{2i-2}{-2i-2}\Rightarrow z=\frac{1-i}{1+i}$
$\Rightarrow z=\frac{1-i}{1+i}\times \frac{1-i}{1-i}\Rightarrow z=\frac{1+i^2-2i}{1-i^2}$
$\Rightarrow z=-i$
$\therefore a=0,b=-1$
$\Rightarrow a\cdot b=0$