Question:
If $z=\frac{1}{(2+3 i)^{2}}$, than $|z|=$
(a) $\frac{1}{13}$
(b) $\frac{1}{5}$
(c) $\frac{1}{12}$
(d) none of these
Solution:
(a) $\frac{1}{13}$
Let $z=\frac{1}{(2+3 i)^{2}}$
$\Rightarrow z=\frac{1}{4+9 i^{2}+12 i}$
$\Rightarrow z=\frac{1}{4-9+12 i}$
$\Rightarrow z=\frac{1}{-5+12 i}$
$\Rightarrow z=\frac{1}{-5+12 i} \times \frac{-5-12 i}{-5-12 i}$
$\Rightarrow z=\frac{-5-12 i}{25+144}$
$\Rightarrow z=\frac{-5}{169}-\frac{12 i}{169}$
$\Rightarrow|z|=\sqrt{\frac{25}{169^{2}}+\frac{144}{169^{2}}}$
$\Rightarrow|z|=\frac{1}{\sqrt{169}}$
$\Rightarrow|z|=\frac{1}{13}$