Question

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

• A. $|z|=2$
• B. $|z|=\frac{1}{2}$
• C. $\text{amp (z)}=\frac{\pi }{4}$
• D. $\text{amp (z)}=\frac{3\pi }{4}$

Answer 1

The correct option is D

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

$z=\frac{1+7i}{(2-i{)}^2}\Rightarrow z=\frac{1+7i}{4+i^2-4i}\Rightarrow z=\frac{1+7i}{4-1-4i}[\because i^2=-1]\Rightarrow z=\frac{1+7i}{3-4i}\Rightarrow z=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}\Rightarrow z=\frac{3+4i+21i+28i^2}{9-16i^2}\Rightarrow z=\frac{3-28+25i}{9+16}\Rightarrow z=\frac{-25+25i}{25}\Rightarrow z=-1+itan\alpha =\left|\frac{Im\left(z\right)}{Re\left(z\right)}\right|=1\Rightarrow \alpha =\frac{\pi }{4}$

Since, z lies in the second quadrant.

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

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