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Convert the given complex number in polar form :$\;-3$

$(a)\;0\qquad(b)\;3[cos (\large\frac{\pi}{2}+i\; sin(\large\frac{\pi}{2})]\qquad(c)\;3[cos \pi + i\; sin \pi]\qquad(d)\;3[cos (-\large\frac{\pi}{2})+i\; sin (-\large\frac{\pi}{2})]$

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Answer : $\;3[cos \pi + i sin \pi]$
Explanation :
$\;-3\;$ Let $\;r cos \theta =-3 \;$ and $\;r sin \theta = 0$
on squaring and adding , we obtain
$(r cos \theta)^{2}+(r sin \theta)^{2} = (-3)^{2}+(0)^{2}$
$r^{2}(cos^{2} \theta + sin^{2} \theta) = 9$
$r^{2} = 9 \qquad [cos^{2} \theta + sin^{2} \theta=1]$
$r =\sqrt{9}=3 \qquad [conventionally , r > 0 ]$
Therefore , $\;3 cos \theta = -3 \;$ and $\; 3 sin \theta = 0$
$\; cos \theta = -1 \;$ and $\; sin \theta = 0$
$\theta = \pi$
$-3 = r cos \theta + i r sin \theta$
$=3[cos \pi + i \;sin \pi]\;$ This is the required polar form
answered Apr 10, 2014 by yamini.v
 
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