Question

Which of the following is a fourth root of $\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$ ?
(A) $cis\left( {\dfrac{\pi }{{12}}} \right)$
(B) $cis\left( {\dfrac{\pi }{2}} \right)$
(C) $cis\left( {\dfrac{\pi }{6}} \right)$
(D) $cis\left( {\dfrac{\pi }{3}} \right)$

Answer 1

Hint: Start with converting the complex number in polar form. And then apply De Moivre’s theorem to take its fourth root. Finally convert the resulting complex number in $cis\theta $ notation to get it in the desired form.

Complete Step-by-Step solution:
The given complex number is $\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$. Let it be z. So, we have:
$ \Rightarrow z = \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}$
We know that modulus of any complex number $x + iy$ is $\sqrt {{x^2} + {y^2}} $. So, modulus of z is:
$
   \Rightarrow \left| z \right| = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}} , \\
   \Rightarrow \left| z \right| = \sqrt {\dfrac{1}{2} + \dfrac{3}{4}} , \\
   \Rightarrow \left| z \right| = 1 \\
$
Similarly, the argument of the complex number is ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$. So, argument of z is:
$
   \Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}} \right), \\
   \Rightarrow \theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right), \\
   \Rightarrow \theta = \dfrac{\pi }{3} \\
$
Thus the polar form of z is:
$
   \Rightarrow z = \left| z \right|\left( {\cos \theta + i\sin \theta } \right), \\
   \Rightarrow z = \cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3} .....(i) \\
$
z can also be written as:
$ \Rightarrow z = cis\dfrac{\pi }{3}$
We have to find out the fourth root of z. So, we have:
\[ \Rightarrow {z^{\dfrac{1}{4}}} = {\left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)^{\dfrac{1}{4}}}{\text{ }}.....(i)\]
According to De Moivre’s theorem:
\[ \Rightarrow {\left( {\cos \theta + i\sin \theta } \right)^n} = \cos n\theta + i\sin n\theta \]
Using this theorem in equation $(i)$, we’ll get:
\[
   \Rightarrow {z^{\dfrac{1}{4}}} = \cos \dfrac{\pi }{{3 \times 4}} + i\sin \dfrac{\pi }{{3 \times 4}}, \\
   \Rightarrow {z^{\dfrac{1}{4}}} = \cos \dfrac{\pi }{{12}} + i\sin \dfrac{\pi }{{12}}, \\
   \Rightarrow {z^{\dfrac{1}{4}}} = cis\left( {\dfrac{\pi }{{12}}} \right) \\
\]
Thus, the fourth root of the complex number is \[cis\left( {\dfrac{\pi }{{12}}} \right)\]. (A) is the correct option.
Note: The complex number \[\left( {\cos \theta + i\sin \theta } \right)\] can also be written as $r{e^{i\theta }}$, where r is the modulus of the complex number. Using this for the above complex number:
\[
   \Rightarrow z = {e^{i\dfrac{\pi }{3}}}, \\
   \Rightarrow {z^{\dfrac{1}{4}}} = {\left( {{e^{i\dfrac{\pi }{3}}}} \right)^{\dfrac{1}{4}}}, \\
   \Rightarrow {z^{\dfrac{1}{4}}} = {e^{i\dfrac{\pi }{{12}}}} = cis\left( {\dfrac{\pi }{{12}}} \right) \\
\]
We used different methods but the end result is the same.