1+7 i (2-i)^2 = - Vidyadip

MCQ

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

✓
$\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
B
$\sqrt{2}\left(\cos \frac{\pi}{4}+ i \sin \frac{\pi}{4}\right)$
C
$\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
D
$\left(\cos \frac{\pi}{4}- i \sin \frac{\pi}{4}\right)$

✓

Answer

Correct option: A.

$\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

(A)
$\frac{1+7 i}{(2-i)^2}=\frac{(1+7 i)}{(3-4 i)} \frac{(3+4 i)}{(3+4 i)}=\frac{-25+25 i}{25}=-1+i$
Let $z =x| i y=-1| i$
$\therefore \quad r \cos \theta=-1$ and $r \sin \theta=1$
$\therefore \quad \theta=\frac{3 \pi}{4}$ and $r=\sqrt{2}$
Thus, $\frac{1+7 i}{(2-i)^2}=\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$

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