Question

How do you find all the real and complex roots of ${x^6} - 64 = 0$ ?

Answer 1

Hint: In the given question, we are required to find all the roots of the equation ${x^6} - 64 = 0$. The roots of the given equation may be real or may be complex. Such questions can also be solved using Demoivre's theorem as it also involves calculating the rational power of a complex number with ease.

Complete step by step answer:
So, we have, ${x^6} - 64 = 0$.
We need to find all the real and complex roots of the given equation. Hence, we have to simplify the equation so as to find the value of x.
${x^6} - 64 = 0$
Shifting $64$ from left side of the equation to right side of the equation, we get,
$ \Rightarrow {x^6} = 64$
We know that $64$can also be written as $64\left( {\cos \left( {2k\pi + 0} \right) + i\sin \left( {2k\pi + 0} \right)} \right)$ in polar form as a complex number because $\cos 0 = 1$and$\sin 0 = 0$. So, writing $64$ as $64\left( {\cos \left( {2k\pi } \right) + i\sin \left( {2k\pi } \right)} \right)$, we get the equation as:
\[ \Rightarrow {x^6} = 64\left( {\cos \left( {2k\pi } \right) + i\sin \left( {2k\pi } \right)} \right)\]
$ \Rightarrow x = {\left[ {64\left( {\cos \left( {2k\pi } \right) + i\sin \left( {2k\pi } \right)} \right)} \right]^{\dfrac{1}{6}}}$
We know that $64 = {2^6}$. So, we get,
$ \Rightarrow x = {\left[ {{2^6}\left( {\cos \left( {2k\pi } \right) + i\sin \left( {2k\pi } \right)} \right)} \right]^{\dfrac{1}{6}}}$
Simplifying further, we get,
$ \Rightarrow x = 2{\left( {\cos \left( {2k\pi } \right) + i\sin \left( {2k\pi } \right)} \right)^{\dfrac{1}{6}}}$
Now, according to Demoivre’s Theorem, we know that \[{\left( {\cos \left( x \right) + i\sin \left( x \right)} \right)^{\dfrac{1}{n}}} = \left( {\cos \left( {\dfrac{x}{n}} \right) + i\sin \left( {\dfrac{x}{n}} \right)} \right)\] for $k = 0,1,2,3,4,5$
So, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2k\pi }}{6}} \right) + i\sin \left( {\dfrac{{2k\pi }}{6}} \right)} \right)$ for $k = 0,1,2,3,4,5$
For $k = 0$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 0 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 0 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \left( 0 \right) + i\sin \left( 0 \right)} \right)$
$ \Rightarrow x = 2\left( 1 \right) = 2$
For $k = 1$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 1 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 1 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{\pi }{3}} \right) + i\sin \left( {\dfrac{\pi }{3}} \right)} \right)$
$ \Rightarrow x = 2\left( {\dfrac{1}{2} + i\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right)$
$ \Rightarrow x = 1 + \sqrt 3 i$
For $k = 2$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 2 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 2 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\pi }}{3}} \right) + i\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right)$
$ \Rightarrow x = 2\left( {\dfrac{{ - 1}}{2} + i\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right)$
$ \Rightarrow x = - 1 + \sqrt 3 i$
For $k = 3$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 3 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 3 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \pi + i\sin \pi } \right)$
$ \Rightarrow x = 2\left( { - 1 + i\left( 0 \right)} \right)$
$ \Rightarrow x = - 2$
For $k = 4$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 4 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 4 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \dfrac{{4\pi }}{3} + i\sin \dfrac{{4\pi }}{3}} \right)$
$ \Rightarrow x = 2\left( { - \dfrac{1}{2} + i\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right)$
$ \Rightarrow x = - 1 - i\sqrt 3 $
For $k = 5$, we get,
$ \Rightarrow x = 2\left( {\cos \left( {\dfrac{{2\left( 5 \right)\pi }}{6}} \right) + i\sin \left( {\dfrac{{2\left( 5 \right)\pi }}{6}} \right)} \right)$
Simplifying further, we get,
$ \Rightarrow x = 2\left( {\cos \dfrac{{5\pi }}{3} + i\sin \dfrac{{5\pi }}{3}} \right)$
$ \Rightarrow x = 2\left( {\dfrac{1}{2} + i\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right)$
$ \Rightarrow x = 1 - i\sqrt 3 $
So, all the real roots of the equation ${x^6} - 64 = 0$ are: $2$ and $ - 2$
Also, all the complex roots of the equation ${x^6} - 64 = 0$ are: $1 - i\sqrt 3 $,$ - 1 - i\sqrt 3 $,$1 + i\sqrt 3 $ and $ - 1 + i\sqrt 3 $.

So, all the roots of the equation ${x^6} - 64 = 0$are: $2$, $ - 2$, $1 - i\sqrt 3 $,$ - 1 - i\sqrt 3 $,$1 + i\sqrt 3 $ and $ - 1 + i\sqrt 3 $.

Note: The application of demoivre’s theorem makes the question a lot easier than the conventional method of finding the power of the complex number. Trigonometry is of much use when dealing with polar forms of complex numbers.