Question

How do you get the complex cube root of $8$?

Answer 1

Hint: This sum is based on the De Moivre's theorem. It is essential that the student knows about this theorem and its application. Even if the student knows about the formula, it is important that the student is acquainted with the concept of De Moivre's theorem. It shows that there are always $3$ cube roots which are placed evenly around a circle. In this particular sum the student has to use the formula ${z^{\dfrac{1}{n}}} = |{z^{\dfrac{1}{n}}}| \times (\cos (\dfrac{{\Phi + 2k\pi }}{n}) + i\sin (\dfrac{{\Phi + 2k\pi }}{n}))$for $k \in \{ 1,2,3..........n - 1\} $. Using this formula the student has to put the value of $n = 3$,since we have to find the cube root and $k = 1$, considering the minimum value.

Complete step by step solution:
Since we are applying the De Moivre's theorem, we must first find the value of its variables before proceeding on to the main sum. Formula of De Moivre's theorem is given as follows:
${z^{\dfrac{1}{n}}} = |{z^{\dfrac{1}{n}}}| \times (\cos (\dfrac{{\Phi + 2k\pi }}{n}) + i\sin (\dfrac{{\Phi + 2k\pi }}{n}))$for $k \in \{ 1,2,3..........n - 1\} $
From the above formula, we have
$
  z = 8 \\
  n = 3 \\
  \Phi = 0 \\
  k = 1 \\
$
Substituting the values in the formula,
${8^{\dfrac{1}{3}}} = |{8^{\dfrac{1}{3}}}| \times (\cos (\dfrac{{2\pi }}{3}) + i\sin (\dfrac{{2\pi }}{3}))...........(1)$
The cube roots of $8$ are plotted in the complex plane on the circle of radius $2$. Equation $1$can be written as
$ \Rightarrow 2 = 2 \times (\cos (\dfrac{{2\pi }}{3}) + i\sin (\dfrac{{2\pi }}{3}))...........(2)$
Since we do not have the direct formula for $\cos (\dfrac{{2\pi }}{3})\& \sin (\dfrac{{2\pi }}{3})$, we will have to use the property
$\cos (\pi - \theta ) = - \cos \theta $, & $\sin (\pi - \theta ) = \sin \theta $,
Equation $2$can be rearranged as
$ \Rightarrow 2 = 2 \times (\cos (\pi - \dfrac{\pi }{3}) + i\sin (\pi - \dfrac{\pi }{3}))...........(3)$
Now we know the values of $\cos (\dfrac{\pi }{3})\& \sin (\dfrac{\pi }{3})$, substituting them in the above equation-
$ \Rightarrow 2 = 2 \times (\cos (\pi - \dfrac{\pi }{3}) + i\sin (\pi - \dfrac{\pi }{3}))...........(3)$
$ \Rightarrow 2 \times ( - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2})...........(4)$
\[ \Rightarrow - 1 + i\sqrt 3 ..........(5)\]
We know that one of the roots of $8$ is $2$, we now have $2$ roots. In order to find the $3rd$ root, we will substitute the value of $k = 2$ ,rest everything remains the same in the main formula. The equation now becomes
${8^{\dfrac{1}{3}}} = |{8^{\dfrac{1}{3}}}| \times (\cos (\dfrac{{4\pi }}{3}) + i\sin (\dfrac{{4\pi }}{3}))...........(6)$
On solving equation $6$as above and re-substituting $\dfrac{{4\pi }}{3}$as$\pi + \dfrac{\pi }{3}$, using the properties of trigonometric functions, we get the third root which is
\[ \Rightarrow 1 - i\sqrt 3 ..........(7)\]
Also we know that $\omega = - \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}i$, is the primitive complex cube root of $1$.
Squaring the above value of $\omega $, we get ${\omega ^2} = - \dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}i$

Thus we can say that complex cube roots of $8$ are $2,2\omega ,2{\omega ^2}$.

Note: Though DeMoivre's Theorem looks complicated to solve, it proves beneficial while solving Laplace equations. Another method to solve this sum is by a quadratic equation. Wherein the student has to find the roots by determining factors of the equation ${x^3} - 8$. And use the formula of the quadratic equation to solve this particular sum. It is always advisable to solve sums of such categories using the above method, as some of the numerical involving complex numbers cannot be solved by using the quadratic equation formula.