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Question: How do you find the solution to \( {x^5} + 1 = 0 \)...

How do you find the solution to x5+1=0{x^5} + 1 = 0

Explanation

Solution

Hint : In order to determine the value of the above question, rewrite the above equation as x5=−15=(−1)15{x^5} = \sqrt[5]{{ - 1}} = {\left( { - 1} \right)^{\dfrac{1}{5}}} and then use the De Moivre’s Theorem to write (−1)15=cos⁡(2nπ+π5)+isin⁡(2nπ+π5){\left( { - 1} \right)^{\dfrac{1}{5}}} = \cos \left( {\dfrac{{2n\pi + \pi }}{5}} \right) + i\sin \left( {\dfrac{{2n\pi + \pi }}{5}} \right) and put value of n from 0 to 4 to determine the roots.

Complete step-by-step answer :
We are given a expression x5+1=0{x^5} + 1 = 0 let it be f(x)f(x)
Equation can be rewritten as
⇒x5=−1 ⇒x5=−15=(−1)15   \Rightarrow {x^5} = - 1 \\\ \Rightarrow {x^5} = \sqrt[5]{{ - 1}} = {\left( { - 1} \right)^{\dfrac{1}{5}}} \;
Consider the fact that −1=cos⁡(π)+isin(π) - 1 = \cos \left( \pi \right) + isin\left( \pi \right)and we can also write this as
−1=cos⁡(2nπ+π)+isin(2nπ+π)- 1 = \cos \left( {2n\pi + \pi } \right) + isin\left( {2n\pi + \pi } \right)
Now using De Moivre’s Theorem,
We can write it as
(−1)15=cos⁡(2nπ+π5)+isin⁡(2nπ+π5){\left( { - 1} \right)^{\dfrac{1}{5}}} = \cos \left( {\dfrac{{2n\pi + \pi }}{5}} \right) + i\sin \left( {\dfrac{{2n\pi + \pi }}{5}} \right) ---------(1)
Since this above equation will have 5 roots which are the solution to equation x5+1=0{x^5} + 1 = 0 and can be obtained by putting n=0,1,2,3n = 0,1,2,3 and 44
So these values are
When n=0n = 0 ,in equation (1)
cos⁡(π5)+isin⁡(π5)\cos \left( {\dfrac{\pi }{5}} \right) + i\sin \left( {\dfrac{\pi }{5}} \right)
When n=1n = 1 ,in equation (1)
cos⁡(3π5)+isin⁡(3π5)=−cos⁡(2π5)+isin⁡(2π5)\cos \left( {\dfrac{{3\pi }}{5}} \right) + i\sin \left( {\dfrac{{3\pi }}{5}} \right) = - \cos \left( {\dfrac{{2\pi }}{5}} \right) + i\sin \left( {\dfrac{{2\pi }}{5}} \right)
When n=2n = 2 ,in equation (1)
cos⁡(5π5)+isin⁡(5π5)=cos⁡π+isin⁡π=−1\cos \left( {\dfrac{{5\pi }}{5}} \right) + i\sin \left( {\dfrac{{5\pi }}{5}} \right) = \cos \pi + i\sin \pi = - 1
When n=3n = 3 ,in equation (1)
cos⁡(7π5)+isin⁡(7π5)=−cos⁡(2π5)−isin⁡(2π5)\cos \left( {\dfrac{{7\pi }}{5}} \right) + i\sin \left( {\dfrac{{7\pi }}{5}} \right) = - \cos \left( {\dfrac{{2\pi }}{5}} \right) - i\sin \left( {\dfrac{{2\pi }}{5}} \right)
When n=4n = 4 ,in equation (1)
cos⁡(9π5)+isin⁡(9π5)=cos⁡(π5)−isin⁡(π5)\cos \left( {\dfrac{{9\pi }}{5}} \right) + i\sin \left( {\dfrac{{9\pi }}{5}} \right) = \cos \left( {\dfrac{\pi }{5}} \right) - i\sin \left( {\dfrac{\pi }{5}} \right)
Therefore the roots to x5+1=0{x^5} + 1 = 0 are cos⁡(π5)−isin⁡(π5)\cos \left( {\dfrac{\pi }{5}} \right) - i\sin \left( {\dfrac{\pi }{5}} \right),−cos⁡(2π5)−isin⁡(2π5) - \cos \left( {\dfrac{{2\pi }}{5}} \right) - i\sin \left( {\dfrac{{2\pi }}{5}} \right),cos⁡π+isin⁡π=−1\cos \pi + i\sin \pi = - 1,−cos⁡(2π5)+isin⁡(2π5) - \cos \left( {\dfrac{{2\pi }}{5}} \right) + i\sin \left( {\dfrac{{2\pi }}{5}} \right)andcos⁡(π5)+isin⁡(π5)\cos \left( {\dfrac{\pi }{5}} \right) + i\sin \left( {\dfrac{\pi }{5}} \right).

Note : .In the field of complex numbers, DeMoivre's Theorem is perhaps the most significant and helpful hypothesis which associates complex numbers and trigonometry. Likewise accommodating for acquiring connections between trigonometric functions of multiple angles. DeMoivre's Theorem otherwise called "De Moivre's Identity" and "De Moivre's Formula". The name of the hypothesis is after the name of extraordinary mathematician De Moivre, who made numerous commitments to the field of mathematics, predominantly in the areas of theory of probability and algebra.