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

$(a)\;0\qquad(b)\;cos (\large\frac{\pi}{6})+i\;sin(\large\frac{\pi}{6})\qquad(c)\;cos (\large\frac{\pi}{4})+i\;sin(\large\frac{\pi}{4})\qquad(d)\;cos (\large\frac{\pi}{2})+i\;sin(\large\frac{\pi}{2})$

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