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Question
\[\text { If } z = \frac{1}{(2 + 3i )^2}, \text { than } \left| z \right| =\]
Options
\[\frac{1}{13}\]
\[\frac{1}{5}\]
\[\frac{1}{12}\]
none of these
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Solution
\[\frac{1}{13}\]
\[\text { Let } z = \frac{1}{\left( 2 + 3i \right)^2}\]
\[ \Rightarrow z = \frac{1}{4 + 9 i^2 + 12i} \]
\[ \Rightarrow z = \frac{1}{4 - 9 + 12i} \]
\[ \Rightarrow z = \frac{1}{- 5 + 12i}\]
\[\Rightarrow z=\frac{1}{- 5 + 12i}\times\frac{- 5 - 12i}{- 5 - 12i}\]
\[\Rightarrow z=\frac{- 5 - 12i}{25 + 144}\]
\[ \Rightarrow z=\frac{- 5}{169}-\frac{12i}{169}\]
\[\Rightarrow\left| z \right|=\sqrt{\frac{25}{{169}^2} + \frac{144}{{169}^2}}\]
\[\Rightarrow \left| z \right|=\frac{1}{\sqrt{169}}\]
\[\Rightarrow \left| z \right| = \frac{1}{13}\]
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