MCQ
If $\frac{1+7\text{i}}{(2-\text{i})^2},$ then:
- A$|\text{z}|=2$
- B$|\text{z}|=\frac{1}{2}$
- C$\text{amp(z)}=\frac{\pi}{4}$
- D$\text{amp(z)}=\frac{3\pi}{4}$
Solution:
$\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{4-1-4\text{i}} \ [\because\text{i}^2=-1]$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}$
$\Rightarrow\text{z}=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}$
$\Rightarrow\text{z}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{9-16\text{i}^2}$
$\Rightarrow\text{z}=\frac{3-28+25\text{i}}{9+16}$
$\Rightarrow\text{z}=\frac{-25+25\text{i}}{25}$
$\Rightarrow\text{z}=-1+\text{i}$
$\Rightarrow\tan\alpha=\Big|\frac{\text{Im(z)}}{\text{Re(z)}}\Big|$
$=1$
$\Rightarrow\alpha=\frac{\pi}{4}$
Since, z lies in the second quadrant.
Therefore, $\text{amp(z)}=\pi-\alpha$
$=\pi-\frac{\pi}{4}$
$=\frac{3\pi}{4}$
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