Question

How do you solve `x^5+1=0`?

Answer 1

Solutions are `cos((pi)/5)+isin((pi)/5)`
`-cos((2pi)/5)+isin((2pi)/5)`, `-1`
`-cos((2pi)/5)-isin((2pi)/5)` and `cos((pi)/5)-isin((pi)/5)`

Explanation:

As `x^5+1=0`, we have `x^5=-1` and `x=root(5)(-1)=(-1)^(1/5)`

Hence solution of `x^5+1=0` means to find fifth roots of `-1`.

Note that as `-1=cospi+isinpi`, and we can also write

`-1=cos(2npi+pi)+isin(2npi+pi)`

and using De Moivre's Theorem

`(-1)^(1/5)=cos((2npi+pi)/5)+isin((2npi+pi)/5)`

and five roots, which are solutions of `x^5+1=0` can be obtained by putting `n=0,1,2,3` and `4` (after `4` roots will start repeating) and these are

`cos((pi)/5)+isin((pi)/5)`
`cos((3pi)/5)+isin((3pi)/5)=-cos((2pi)/5)+isin((2pi)/5)`
`cos((5pi)/5)+isin((5pi)/5)=cospi+isinpi=-1`
`cos((7pi)/5)+isin((7pi)/5)=-cos((2pi)/5)-isin((2pi)/5)`
`cos((9pi)/5)+isin((9pi)/5)=cos((pi)/5)-isin((pi)/5)`