Convert the given complex number in polar for | vedclass

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Question

(A) Let the complex number be $z = i = 0 + 1i$.The polar form is given by $z = r(\cos 	heta + i \sin 	heta)$,where $r \cos 	heta = 0$ and $r \sin \

Vedclass · Accepted Answer

$\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$

Vedclass · Answer

(A) Let the complex number be $z = i = 0 + 1i$.The polar form is given by $z = r(\cos 	heta + i \sin 	heta)$,where $r \cos 	heta = 0$ and $r \sin 	heta = 1$.Squaring and adding both equations:$r^2(\cos^2 	heta + \sin^2 	heta) = 0^2 + 1^2$$r^2(1) = 1$$r = 1$ (since $r > 0$).Now,$\cos 	heta = 0$ and $\sin 	heta = 1$.This implies $	heta = \frac{\pi}{2}$.Therefore,the polar form is $1(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$.

List-$I$ (Complex number)	List-$II$ (Polar form)
$(i) \sqrt{3}-i$	$(a) 2 \operatorname{cis} \frac{\pi}{6}$
$(ii) \sqrt{3}+i$	$(b) 2 \operatorname{cis} \frac{5 \pi}{6}$
$(iii) -\sqrt{3}+i$	$(c) 2 \operatorname{cis}\left(-\frac{5 \pi}{6}\right)$
$(iv) -\sqrt{3}-i$	$(d) 2 \operatorname{cis}\left(-\frac{\pi}{6}\right)$

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

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