Question

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

• A. $cis\frac{\pi }{12}$
• B. $cis\frac{\pi }{2}$
• C. $cis\frac{\pi }{6}$
• D. $cis\frac{\pi }{3}$

Answer 1

The correct option is A

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

Explanation for the correct option:

Step 1: Change the equation into polar form

Complex 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 $z=r(cos\theta +isin\theta )$ ….(2)

$r=\sqrt{a^2+b^2}$

$$\begin{gathered} \Rightarrow r=\sqrt{(\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2} \\ \Rightarrow r=\sqrt{\frac{1}{4}+\frac{3}{4}} \\ \Rightarrow r=\sqrt{\frac{1+3}{4}} \\ \Rightarrow r=\sqrt{\frac{4}{4}} \\ \Rightarrow r=\sqrt{1} \\ \Rightarrow r=1 \end{gathered}$$

Now

$$\begin{gathered} cos\theta =\frac{a}{r} \\ \Rightarrow cos\theta =\frac{1}{2} \\ \Rightarrow cos\theta =cos\frac{\pi }{3} \end{gathered}$$

and

$$\begin{gathered} sin\theta =\frac{b}{r} \\ \Rightarrow sin\theta =\frac{\sqrt{3}}{2} \\ \Rightarrow sin\theta =sin\frac{\pi }{3} \end{gathered}$$

Equate equation(1) and equation(2),

$$\begin{gathered} \frac{1}{2}+i\frac{\sqrt{3}}{2}=r(cos\theta +isin\theta ) \\ \Rightarrow \frac{1}{2}+i\frac{\sqrt{3}}{2}=cos\frac{\pi }{3}+isin\frac{\pi }{3} \end{gathered}$$

Step 2: Write the equation in exponential form

The exponential form of a complex number $\cos \theta +i\sin \theta$ is $e^{i\theta }$.

$cos\frac{\pi }{3}+isin\frac{\pi }{3}=e^{\frac{i\pi }{3}}$

The fourth root of the equation is,

$$\begin{gathered} \sqrt[4]{e^{\frac{i\pi }{3}}}=e^{\frac{i\pi }{12}} \\ {e}^{\frac{i\pi }{12}}=cos\frac{\pi }{12}+isin\frac{\pi }{12} \end{gathered}$$

Step 3: Write the equation in cis from

We know that, $cosx+isinx=cisx$

$cos\frac{\pi }{12}+isin\frac{\pi }{12}=cis\frac{\pi }{12}$

Hence, the correct option is option (a).