Match the items of List-I with those of List- | vedclass

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(B) For a complex number $z = x + iy$,the polar form is $r \operatorname{cis} 	heta$,where $r = \sqrt{x^2 + y^2}$ and $	heta = \operatorname{arg}(z)

Vedclass · Accepted Answer

$(i)-d, (ii)-a, (iii)-b, (iv)-c$

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(B) For a complex number $z = x + iy$,the polar form is $r \operatorname{cis} 	heta$,where $r = \sqrt{x^2 + y^2}$ and $	heta = \operatorname{arg}(z)$.$(i) z = \sqrt{3} - i$: $r = \sqrt{3+1} = 2$. $	heta = 	an^{-1}(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}$. Thus,$z = 2 \operatorname{cis}(-\frac{\pi}{6})$ (Matches $d$).$(ii) z = \sqrt{3} + i$: $r = 2$. $	heta = 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{\pi}{6}$. Thus,$z = 2 \operatorname{cis}(\frac{\pi}{6})$ (Matches $a$).$(iii) z = -\sqrt{3} + i$: $r = 2$. $	heta = \pi - 	an^{-1}(\frac{1}{\sqrt{3}}) = \frac{5\pi}{6}$. Thus,$z = 2 \operatorname{cis}(\frac{5\pi}{6})$ (Matches $b$).$(iv) z = -\sqrt{3} - i$: $r = 2$. $	heta = -\pi + 	an^{-1}(\frac{1}{\sqrt{3}}) = -\frac{5\pi}{6}$. Thus,$z = 2 \operatorname{cis}(-\frac{5\pi}{6})$ (Matches $c$).Therefore,the correct matching is $(i)-d, (ii)-a, (iii)-b, (iv)-c$.

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

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