Solve the following


If $z=\frac{1+7 i}{(2-i)^{2}}$, then

(a) $|z|=2$

(b) $|z|=\frac{1}{2}$

(c) $\operatorname{amp}(z)=\frac{\pi}{4}$

(d) $\operatorname{amp}(z)=\frac{3 \pi}{4}$


(d) $\operatorname{amp}(z)=\frac{3 \pi}{4}$

$z=\frac{1+7 i}{(2-i)^{2}}$

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

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

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

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

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

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

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

$\Rightarrow z=-1+i$

$\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

= 1

$\Rightarrow \alpha=\frac{\pi}{4}$

Since, $z$ lies in the second quadrant.

Therefore, $\operatorname{amp}(z)=\pi-\alpha$


$=\frac{3 \pi}{4}$

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