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Convert the given complex number in polar form :$\;-1-i\;$

$(a)\;\sqrt{2}\;[cos (\large\frac{\pi}{2})+i \;sin(\large\frac{\pi}{2})]\qquad(b)\;\sqrt{2}\;[cos (-\large\frac{3\pi}{4})+i \;sin(-\large\frac{3\pi}{4})]\qquad(c)\;\sqrt{2}\;[cos (-\large\frac{\pi}{2})+i \;sin(-\large\frac{\pi}{2})]\qquad(d)\;\sqrt{2}\;[cos (\large\frac{3\pi}{2})+i \;sin(\large\frac{3\pi}{2})]$

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Answer : $\;\sqrt{2}\;[cos (-\large\frac{3\pi}{4})+i \;sin(-\large\frac{3\pi}{4})]$
Explanation :
$\;-1-i\;$ Let $\;r cos \theta =-1 \;$ and $\;r sin \theta = -1$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-1)^{2}+(-1)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 1+ 1$
$r^{2} = 2 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{2} \qquad [conventionally , r > 0 ]$
Therefore , $\; \sqrt{2} cos \theta =- 1 \;$ and $\; \sqrt{2} sin \theta = - 1$
$cos \theta = \large\frac{-1}{\sqrt{2}}\;$ and $\;sin \theta = \large\frac{-1}{\sqrt{2}}$
$\theta = - (\pi-\large\frac{\pi}{4})=-\large\frac{3\pi}{4}\qquad [\theta \;lies\; in\;the\;3^{rd} \; quadrant]$
$-1-i = r cos \theta + i r sin \theta$
$=\sqrt{2}\;[cos (-\large\frac{3\pi}{4})+i \;sin(-\large\frac{3\pi}{4})]\;$ this is the required polar form .
answered Apr 10, 2014 by yamini.v
 
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