Question

$\frac{1+2i+3i^2}{1-2i+3i^2}$ equals
(a) i
(b) −1
(c) −i
(d) 4

Answer 1

(c) $-$i

$$\begin{gathered} \mathrm{Let}z=\frac{1+2i+3i^2}{1-2i+3i^2} \\ \Rightarrow z=\frac{1+2i-3}{1-2i-3} \\ \Rightarrow z=\frac{-2+2i}{-2-2i}\times \frac{-2+2i}{-2+2i} \\ \Rightarrow z=\frac{{\left(-2+2i\right)}^2}{{\left(-2\right)}^2-{\left(2i\right)}^2} \\ \Rightarrow z=\frac{4+4i^2-8i}{4+4} \\ \Rightarrow z=\frac{4-4-8i}{8} \\ \Rightarrow z=\frac{-8i}{8} \\ \Rightarrow z=-i \end{gathered}$$