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) = \[\frac{3\pi}{4}\]

Answer 1

amp (z) = \[\frac{3\pi}{4}\]

\[z = \frac{1 + 7i}{\left( 2 - i \right)^2}\]

\[ \Rightarrow z = \frac{1 + 7i}{4 + i^2 - 4i}\]

\[ \Rightarrow z = \frac{1 + 7i}{4 - 1 - 4i} \left[ \because i^2 = - 1 \right]\]

\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i}\]

\[ \Rightarrow z = \frac{1 + 7i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}\]

\[ \Rightarrow z = \frac{3 + 4i + 21i + 28 i^2}{9 - 16 i^2}\]

\[ \Rightarrow z = \frac{3 - 28 + 25i}{9 + 16}\]

\[ \Rightarrow z = \frac{- 25 + 25i}{25}\]

\[ \Rightarrow z = - 1 + i\]

\[\tan \alpha = \left| \frac{Im\left( z \right)}{Re\left( z \right)} \right|\]

\[ = 1\]

\[ \Rightarrow \alpha = \frac{\pi}{4}\]

\[\text { Since, z lies in the second quadrant }. \]

\[\text { Therefore, amp } (z) = \pi - \alpha\]

\[ = \pi - \frac{\pi}{4} \]

\[ = \frac{3\pi}{4} \]