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Question
If \[z = \frac{1 + 7i}{(2 - i )^2}\] , then
Options
\[\left| z \right| = 2\]
\[\left| z \right| = \frac{1}{2}\]
amp (z) = \[\frac{\pi}{4}\]
amp (z) = \[\frac{3\pi}{4}\]
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Solution
amp (z) = \[\frac{3\pi}{4}\]
\[z = \frac{1 + 7i}{\left( 2 - i \right)^2}\]
\[ \Rightarrow z = \frac{1 + 7i}{4 + i^2 - 4i}\]
\[ \Rightarrow z = \frac{1 + 7i}{4 - 1 - 4i} \left[ \because i^2 = - 1 \right]\]
\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i}\]
\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}\]
\[ \Rightarrow z = \frac{3 + 4i + 21i + 28 i^2}{9 - 16 i^2}\]
\[ \Rightarrow z = \frac{3 - 28 + 25i}{9 + 16}\]
\[ \Rightarrow z = \frac{- 25 + 25i}{25}\]
\[ \Rightarrow z = - 1 + i\]
\[\tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]
\[ = 1\]
\[ \Rightarrow \alpha = \frac{\pi}{4}\]
\[\text { Since, z lies in the second quadrant }. \]
\[\text { Therefore, amp } (z) = \pi - \alpha\]
\[ = \pi - \frac{\pi}{4} \]
\[ = \frac{3\pi}{4} \]
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