Question

$\mathrm{If}z=\frac{1}{(2+3i{)}^2},\mathrm{than}\left|z\right|=$
(a) $\frac{1}{13}$
(b) $\frac{1}{5}$
(c) $\frac{1}{12}$
(d) none of these

Answer 1

(a) $\frac{1}{13}$
$$\begin{gathered} \mathrm{Let}z=\frac{1}{{\left(2+3i\right)}^2} \\ \Rightarrow z=\frac{1}{4+9i^2+12i} \\ \Rightarrow z=\frac{1}{4-9+12i} \\ \Rightarrow z=\frac{1}{-5+12i} \end{gathered}$$

$\Rightarrow z=\frac{1}{-5+12i}\times \frac{-5-12i}{-5-12i}$

$$\begin{gathered} \Rightarrow z=\frac{-5-12i}{25+144} \\ \Rightarrow z=\frac{-5}{169}-\frac{12i}{169} \end{gathered}$$

$\Rightarrow \left|z\right|=\sqrt{{\displaystyle \frac{25}{{169}^2}}+{\displaystyle \frac{144}{{169}^2}}}$
$\Rightarrow \left|z\right|={\displaystyle \frac{1}{\sqrt{169}}}$
$\Rightarrow \left|z\right|=\frac{1}{13}$