Question:
$\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$ equals
(a) i
(b) −1
(c) −i
(d) 4
Solution:
(c) −i
Let $z=\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$
$\Rightarrow z=\frac{1+2 i-3}{1-2 i-3}$
$\Rightarrow z=\frac{-2+2 i}{-2-2 i} \times \frac{-2+2 i}{-2+2 i}$
$\Rightarrow z=\frac{(-2+2 i)^{2}}{(-2)^{2}-(2 i)^{2}}$
$\Rightarrow z=\frac{4+4 i^{2}-8 i}{4+4}$
$\Rightarrow z=\frac{4-4-8 i}{8}$
$\Rightarrow z=\frac{-8 i}{8}$
$\Rightarrow z=-i$