Solve the following
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 $z=\frac{1}{(2+3 i)^{2}}$

$\Rightarrow z=\frac{1}{4+9 i^{2}+12 i}$

$\Rightarrow z=\frac{1}{4-9+12 i}$

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

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

$\Rightarrow z=\frac{-5-12 i}{25+144}$

$\Rightarrow z=\frac{-5}{169}-\frac{12 i}{169}$

$\Rightarrow|z|=\sqrt{\frac{25}{169^{2}}+\frac{144}{169^{2}}}$

$\Rightarrow|z|=\frac{1}{\sqrt{169}}$

$\Rightarrow|z|=\frac{1}{13}$

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