Question

Complex number in the form of $A+iB$
$\frac{1+2i+3i^2}{1-2i+3i^2}$

Answer 1

$\left(\frac{1+2i+3i^2}{1-2i+3i^2}\right)$

$=\left(\frac{1+2i-3}{1-2i-3}\right)$

$=\left(\frac{-2+2i}{-2-2i}\right)$

$=\left(\frac{-1+i}{-(1+i)}\right)\left(\frac{1-i}{1-i}\right)$

$=\left(\frac{-(1-i{)}^2}{-(1^2-i^2)}\right)$

$=\left(\frac{1^2+i^2-2i}{1+1}\right)$

$=\left(\frac{-2i}{2}\right)=-i=0+(-1)i$