Question

How do you find all real and complex roots of ${x^5} - 32 = 0$ ?

Answer 1

The given question is pretty easy and solvable. Find the real solution first by using logarithm. Then find complex roots by using Euler’s form of complex numbers. So, let’s have a look at the approach.

Complete Step by Step Solution:
Given that
$\Rightarrow {x^5} = 32$
Taking logarithm on both sides,
$\Rightarrow \log \left( {{x^5}} \right) = \log 32$
Solving further,
$\Rightarrow 5\log x = 5\log 2$
Since logarithm is a one-one function,
Hence $x = 2$.
We obtained the real solution to the equation.
Euler’s form:
$\Rightarrow {e^{i\theta }} = \cos \theta + i\sin \theta$
As you can clearly see,
$\Rightarrow {e^{i \times 2\pi }} = 1$
Because
$\cos 2\pi = 1$ and $\sin 2\pi = 0$
In fact this will be true for all multiples of $2\pi$ .
So we can rewrite 1 as,
$\Rightarrow 1 = {e^{i \times 2\pi }} = {e^{i \times 4\pi }} = {e^{i \times 6\pi }} = {e^{i \times 8\pi }}$
So coming back to our question,
we can write 32 as
$\Rightarrow {x^5} = 32 \times 1$
$\Rightarrow {x^5} = 32{e^{i \times 2\pi }} = 32{e^{i \times 4\pi }} = 32{e^{i \times 6\pi }} = 32{e^{i \times 8\pi }}$
Taking logarithm for all the terms
$\Rightarrow \log {x^5} = \log \left( {32{e^{i \times 2\pi }}} \right) = \log \left( {32{e^{i \times 4\pi }}} \right) = \log \left( {32{e^{i \times 6\pi }}} \right) = \log \left( {32{e^{i \times 8\pi }}} \right)$
$\Rightarrow 5\log x = 5\log \left( {2{e^{i \times \dfrac{{2\pi }}{5}}}} \right) = 5\log \left( {2{e^{i \times \dfrac{{4\pi }}{5}}}} \right) = 5\log \left( {2{e^{i \times \dfrac{{6\pi }}{5}}}} \right) = 5\log \left( {2{e^{i \times \dfrac{{8\pi }}{5}}}} \right)$
Dividing all terms by 5,
$\Rightarrow \log x = \log \left( {2{e^{i \times \dfrac{{2\pi }}{5}}}} \right) = \log \left( {2{e^{i \times \dfrac{{4\pi }}{5}}}} \right) = \log \left( {2{e^{i \times \dfrac{{6\pi }}{5}}}} \right) = \log \left( {2{e^{i \times \dfrac{{8\pi }}{5}}}} \right)$
By this equation you can identify the complex roots for ${x^5} - 32 = 0$.
We have 4 complex roots here,
$\Rightarrow x = 2{e^{i \times \dfrac{{2\pi }}{5}}}$
$\Rightarrow x = 2{e^{i \times \dfrac{{4\pi }}{5}}}$
$\Rightarrow x = 2{e^{i \times \dfrac{{6\pi }}{5}}}$
$\Rightarrow x = 2{e^{i \times \dfrac{{8\pi }}{5}}}$
These all are obtained from Euler's form.
You can easily convert them by using
${e^{i\theta }} = \cos \theta + i\sin \theta$
So,
$\Rightarrow 2{e^{i \times \dfrac{{2\pi }}{5}}} = 2\cos \left( {\dfrac{{2\pi }}{5}} \right) + 2i\sin \left( {\dfrac{{2\pi }}{5}} \right)$
Similarly,
$\Rightarrow 2{e^{i \times \dfrac{{4\pi }}{5}}} = 2\cos \left( {\dfrac{{4\pi }}{5}} \right) + 2i\sin \left( {\dfrac{{4\pi }}{5}} \right)$
$\Rightarrow 2{e^{i \times \dfrac{{6\pi }}{5}}} = 2\cos \left( {\dfrac{{6\pi }}{5}} \right) + 2i\sin \left( {\dfrac{{6\pi }}{5}} \right)$
$\Rightarrow 2{e^{i \times \dfrac{{8\pi }}{5}}} = 2\cos \left( {\dfrac{{8\pi }}{5}} \right) + 2i\sin \left( {\dfrac{{8\pi }}{5}} \right)$
We can substitute some trigonometric terms,
$\because \sin \left( {2\pi - x} \right) = - \sin x$
$\therefore \sin \dfrac{{6\pi }}{5} = - \sin \dfrac{{4\pi }}{5}$
and
$\therefore \sin \dfrac{{8\pi }}{5} = - \sin \dfrac{{2\pi }}{5}$
Similarly for cos,
$\because \cos \left( {2\pi - x} \right) = \cos x$
$\therefore \cos \dfrac{{6\pi }}{5} = \cos \dfrac{{4\pi }}{5}$
and
$\therefore \cos \dfrac{{8\pi }}{5} = \cos \dfrac{{2\pi }}{5}$
Substituting according to the trigonometric transformations,
the complex roots will be
$\Rightarrow x = 2\cos \dfrac{{2\pi }}{5} \pm 2i\sin \dfrac{{2\pi }}{5}$ and
$\Rightarrow x = 2\cos \dfrac{{4\pi }}{5} \pm 2i\sin \dfrac{{4\pi }}{5}$
All the roots of the equation ${x^5} - 32 = 0$ are
$\Rightarrow x = 2$
$\Rightarrow x = 2\cos \dfrac{{2\pi }}{5} \pm 2i\sin \dfrac{{2\pi }}{5}$

$\Rightarrow x = 2\cos \dfrac{{4\pi }}{5} \pm 2i\sin \dfrac{{4\pi }}{5}$

Note:
You could have used other multiples of 2π as well, but all that will lead to these answers at the end. Since the equation is of degree 5, so it will have a total of 5 roots including real and complex roots.