Solve the following


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

(a) i

(b) −1

(c) −i

(d) 4


(c) −i

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

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

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

$\Rightarrow z=\frac{(-2+2 i)^{2}}{(-2)^{2}-(2 i)^{2}}$

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

$\Rightarrow z=\frac{4-4-8 i}{8}$

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

$\Rightarrow z=-i$

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