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What is the polar form of the complex number $\;(i^{25})^{3} \;$ ?

$(a)\; cos \large\frac{\pi}{2} - i sin \large\frac{\pi}{2}\qquad(b)\;cos \large\frac{5\pi}{2} + i sin \large\frac{5\pi}{2}\qquad(c)\;cos \large\frac{3\pi}{2} + i sin \large\frac{3\pi}{2}\qquad(d)\;cos \large\frac{\pi}{4} - i sin \large\frac{\pi}{4}$

1 Answer

Answer : $\; cos \large\frac{\pi}{2} - i sin \large\frac{\pi}{2}$
Explanation :
$\;z=(i^{25})^{3} \; = i^{75}$
$= i^{4 \times 18 +3} = (i^{4})^{18} \;.i^{3}$
$i^{3} =-i = 0-i$
Polar form of $\;z= r(cos \theta + i sin \theta)$
$= 1 \; [cos (- \large\frac{\pi}{2}) + i sin (-\large\frac{\pi}{2})]$
$ =cos \large\frac{\pi}{2} - i sin \large\frac{\pi}{2}$
answered Apr 18, 2014 by yamini.v
 
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