Question

How do you solve `x^4+i=0`?

Answer 1

`= cis ((-pi)/8 ), cis ((3pi)/8 ), cis ((7pi)/8 ), cis ((11pi)/8 )`

Explanation:

`x^4 = - i`

`x = (- i)^(1/4)`

Using Euler's formula (or DeMoivre's theorem, if you like):

`-i = cos ((-pi)/2 + 2n pi) + i sin ((-pi)/2 + 2n pi) equiv cis ((-pi)/2 + 2n pi)`

`= e^(i ((-pi)/2 + 2 n pi))`

So:

`(-i)^(1/4) = e^(i ((-pi)/2 + 2 n pi)/4)`

`= e^(i ((-pi)/8 + (n pi)/2))`

`= cis ((-pi)/8 + (n pi)/2)`

Running through the `n`'s:

`= cis ((-pi)/8 ), cis ((3pi)/8 ), cis ((7pi)/8 ), cis ((11pi)/8 )`

And then they start repeating.

Answer 2

The four roots are:

`x=+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)`

`x=+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)`

Explanation:

Here's an alternative non-trigonometric method...

If `a, b` are real numbers with `b != 0`, then the square roots of `a+bi` are:

`+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)`

(See: https://socratic.org/s/aES256Hm)

Applying this to the square root of `-i = color(blue)(0)+(color(blue)(-1))i` we find that its square roots are:

`+-((sqrt((sqrt(color(blue)(0)^2+(color(blue)(-1))^2)+color(blue)(0))/2)) - (sqrt((sqrt(color(blue)(0)^2+(color(blue)(-1))^2)-color(blue)(0))/2))i)`

`=+-(sqrt(2)/2-sqrt(2)/2i)`

The square roots of `sqrt(2)/2-sqrt(2)/2i` are:

`+-((sqrt((sqrt((color(blue)(sqrt(2)/2))^2+(color(blue)(-sqrt(2)/2))^2)+color(blue)(sqrt(2)/2))/2)) - (sqrt((sqrt((color(blue)(sqrt(2)/2))^2+(color(blue)(-sqrt(2)/2))^2)-color(blue)(sqrt(2)/2))/2))i)`

`=+-((sqrt((sqrt(1/2+1/2)+sqrt(2)/2))/2)) - (sqrt((sqrt(1/2+1/2)-sqrt(2)/2)/2))i)`

`=+-((sqrt((1+sqrt(2)/2)/2)) - (sqrt((1-sqrt(2)/2)/2))i)`

`=+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)`

Rather than simplifying again, note that we can find the square roots of `-sqrt(2)/2+sqrt(2)/2i` by multiplying these by `i` to get:

`+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)`

So the fourth roots of `-i` (i.e. the roots of `x^4+i = 0`) are:

`+-(sqrt(2+sqrt(2))/2 - sqrt(2-sqrt(2))/2i)`

`+-(sqrt(2-sqrt(2))/2 + sqrt(2+sqrt(2))/2i)`