Question

How do you find all the real and complex roots of `x^7 + 128 = 0`?

Answer 1

`x^7+128 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))`

Hence roots:

`x = -2cos((2kpi)/7)-2isin((2kpi)/7)`

where `k=0,1,2,3,4,5,6`

Explanation:

`x^7+128 = x^7+2^7`

By De Moivre's formula we find:

`(cos((2kpi)/7) + i sin((2kpi)/7))^7 = cos(2kpi) + i sin(2kpi) = 1`

for any integer `k`.

So the `7`th roots of `1` can be written as:

`cos((2kpi)/7) + i sin((2kpi)/7)` where `k=0,1,2,3,4,5,6`

Hence we find:

`x^7+2^7 = prod_(k = 0)^6 (x + 2cos((2kpi)/7) + 2isin((2kpi)/7))`

Answer 2

`z = 2cis(pi + (2pik)/7)` for `k = 0,1,2,3,4,5,6`

Explanation:

We have

`x^7 + 128 = 0`

Numerically we know that `-2` is one of the roots.
`z = -2` has modulo `2` and `theta = pi`, so we have the trigonometric representation

`z = 2cis(pi)`

(Where `cis(theta) = cos(theta) + isin(theta)`)

But the thing is, whenever we take the nth root of a number, we bissect the circle of `root(n)(|z|)` radius in `n` parts of equal distance. I.e.: if we take the square roots, the two roots will be exactly `(2pi)/2` degrees apart, if we take the cubic root, they'll be `(2pi)/3` apart and so on.

That means that we can derive that the general form of the roots are

`z = 2cis(pi + (2pik)/7)` for `k = 0,1,2,3,4,5,6` as when we have `k = 7`, we go back to the same root we already had.

Which we can easily check back using De Moivre's formula, since

`z^7 = 128cis(7pi + 7(2pik)/7) = 128cis(7pi + 2pik)`

For any of the mentioned `k`, adding or subtracting `2pi` is the same as doing nothing so

`z^7 = 128cis(7pi)`

And since `7pi = 6pi + pi`

`z^7 = 128cis(pi) = -128`