    Question

# Which of the following is a fourth root of $\frac{1}{2}+i\frac{\sqrt{3}}{2}$?

A

$cis\frac{\pi }{12}$

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B

$cis\frac{\pi }{2}$

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C

$cis\frac{\pi }{6}$

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D

$cis\frac{\pi }{3}$

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Solution

## The correct option is A $cis\frac{\pi }{12}$Explanation for the correct option:Step 1: Change the equation into polar formComplex equation $Z=a+ib$Let $Z=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ ….(1)Therefore, $a=\frac{1}{2}$ and $b=\frac{\sqrt{3}}{2}$Polar form $z=r\left(cos\theta +isin\theta \right)$ ….(2)$r=\sqrt{{a}^{2}+{b}^{2}}$$⇒r=\sqrt{\left(\frac{1}{2}{\right)}^{2}+\left(\frac{\sqrt{3}}{2}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{\frac{1}{4}+\frac{3}{4}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{\frac{1+3}{4}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{\frac{4}{4}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{1}\phantom{\rule{0ex}{0ex}}⇒r=1$Now$cos\theta =\frac{a}{r}\phantom{\rule{0ex}{0ex}}⇒cos\theta =\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒cos\theta =cos\frac{\pi }{3}$and $sin\theta =\frac{b}{r}\phantom{\rule{0ex}{0ex}}⇒sin\theta =\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}⇒sin\theta =sin\frac{\pi }{3}$Equate equation(1) and equation(2),$\frac{1}{2}+i\frac{\sqrt{3}}{2}=r\left(cos\theta +isin\theta \right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}+i\frac{\sqrt{3}}{2}=cos\frac{\pi }{3}+isin\frac{\pi }{3}$Step 2: Write the equation in exponential formThe exponential form of a complex number $\mathrm{cos}\theta +i\mathrm{sin}\theta$ is ${e}^{i\theta }$.$cos\frac{\pi }{3}+isin\frac{\pi }{3}={e}^{\frac{i\pi }{3}}$The fourth root of the equation is,$\sqrt{{e}^{\frac{i\pi }{3}}}={e}^{\frac{i\pi }{12}}\phantom{\rule{0ex}{0ex}}{e}^{\frac{i\pi }{12}}=cos\frac{\pi }{12}+isin\frac{\pi }{12}$Step 3: Write the equation in cis fromWe know that, $cosx+isinx=cisx$$cos\frac{\pi }{12}+isin\frac{\pi }{12}=cis\frac{\pi }{12}$Hence, the correct option is option (a).  Suggest Corrections  1      Similar questions  Explore more