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

MCQ

$\frac{{1 + 7i}}{{{{(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
None of these

✓

Answer

Correct option: A.

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

a
(a) $\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = \frac{{(1 + 7i)}}{{(3 - 4i)}}\frac{{(3 + 4i)}}{{(3 + 4i)}} = \frac{{ - 25 + 25i}}{{25}} = - 1 + i$
Let $z = x + iy = - 1 + i$
$\therefore r\cos \theta = - 1$and $r\sin \theta =1$ $\therefore \theta = \frac{{3\pi }}{4}$and $r = \sqrt 2 $
Thus $\frac{{1 + 7i}}{{{{(2 - i)}^2}}} = \sqrt 2 \left[ {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right]$
Aliter : $\left| {\frac{{1 + 7i}}{{{{(2 - i)}^2}}}} \right| = \left| {\frac{{1 + 7i}}{{3 - 4i}}} \right| = \sqrt 2 $
and $arg\left( {\frac{{1 + 7i}}{{3 - 4i}}} \right) = {\tan ^{ - 1}}7 - {\tan ^{ - 1}}\left( { - \frac{4}{3}} \right)$
$ = {\tan ^{ - 1}}7 + {\tan ^{ - 1}}\frac{4}{3} = \frac{{3\pi }}{4}$
$\therefore \frac{{1 + 7i}}{{{{(2 - i)}^2}}}$$ = \sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right)$

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