Question

The principal mod-amp form of $\frac{1+7i}{(2-i{)}^2}$ is

• A. $\sqrt{5}cis\frac{\pi }{4}$
• B. $\sqrt{2}cis\frac{3\pi }{4}$
• C. $\sqrt{6}cis\frac{\pi }{4}$
• D. $\sqrt{3}cis\frac{3\pi }{4}$

Answer 1

The correct option is B $\sqrt{2}cis\frac{3\pi }{4}$

Assuming $z=a+ib=\frac{1+7i}{(2-i{)}^2}$
we have $(a+ib)(2-i{)}^2=1+7i$
$(a+ib)(3-4i)=1+7i$
$3a+4b+(3b-4a)i=1+7i$
$\therefore$ we have $3a+4b=1$ and $3b-4a=7$
$\therefore$ $12a+16b=4$----------(1)

$9b-12a=21$----------(2)
adding (1) and (2)
$25b=25$
$\therefore b=1$
$\therefore a=-1$
$\therefore$ $\left|z\right|=\sqrt{a^2+b^2}=\sqrt{1+1}=\sqrt{2}$
amp(z)= $\pi -{\tan }^{-1}\left(\left|b/a\right|\right)$ ($\because$ a is -ve,b +ve)
$=\pi -{\tan }^{-1}\left(1\right)$
$=\pi -\pi /4$
$=3\pi /4$