Question

Write down complex numbers in the form of $A+iB$.
$\frac{1+2i+3i^2}{1-2i+3i^2}$.

Answer 1

We have,

$\frac{1+2i+3i^2}{1-2i+3i^2}$

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

$=\frac{1+2i-3}{1-2i-3}$

$=\frac{-2+2i}{-2-2i}$

$=\frac{2-2i}{2+2i}$

$=\frac{2-2i}{2+2i}\times \frac{2-2i}{2-2i}$

$=\frac{4+4i^2-8i}{4-4i^2}$

$=\frac{4-4-8i}{4+4}$

$=\frac{-8i}{8}$

$=0-i$

Hence, this is the answer.