Question

How do you find all solutions to `x^5+243=0`?

Answer 1

`x in {-3e^(i(2pin)/5)| n in {0, 1, 2, 3, 4}}`

Explanation:

We will make use of the fact that `e^(itheta) = e^(i(theta+2pik))` where `k` is an integer (this is clear from Euler's formula `e^(itheta) = cos(theta)+isin(theta))`, along with that `e^0 = 1`.

`x^5+243 = 0`

`=> x^5 = -243 =-243e^(i(0+2pik)), k in ZZ`

`=> x = (-243e^(i2pik))^(1/5) = -3e^(i(2pik)/5)`

Due to the periodic nature of `e^(itheta)`, we have that for any `k in ZZ`, `e^(i(2pik)/5)=e^(i(2pin)/5)` for some `n in {0, 1, 2, 3, 4}`. Thus, we can find our five possible values for `x` by substituting in each possible value for `n`.

`x in {-3e^(i(2pin)/5)| n in {0, 1, 2, 3, 4}}`

Note that as `e^(itheta) = cos(theta)+isin(theta)`, and `sin((2pin)/5) = 0` only when `n = 0`, the only real solution to the equation is `-3e^(i(2pi*0)/5) = -3e^0=-3`

Answer 2

Solutions in `a+bi` form:

`x = -3`

`x = (3(1+sqrt(5)))/4 +- (3sqrt(10 - 2sqrt(5)))/4i`

`x = (3(1-sqrt(5)))/4 +- (3sqrt(10 + 2sqrt(5)))/4i`

Explanation:

Here's an alternative approach to find the solutions in `a+bi` form...

`color(white)`
Fifth roots of `1`

First let us find the Complex fifth roots of `1` in `a+bi` form:

`x^5-1 = (x-1)(x^4+x^3+x^2+x+1)`

`1/(x^2)(x^4+x^3+x^2+x+1) = x^2+x+1+1/x+1/x^2`

`color(white)(1/(x^2)(x^4+x^3+x^2+x+1)) = (x+1/x)^2+(x+1/x)-1`

Using the quadratic formula we find zeros:

`x+1/x = -1/2+-sqrt(5)/2`

Hence:

`x^2+(1/2+-sqrt(5)/2)x+1 = 0`

`color(white)()`
Case `bb(x^2+(1/2+sqrt(5)/2)x+1 = 0)`

Using the quadratic formula:

`x = (-(1/2+sqrt(5)/2) +- sqrt((1/2+sqrt(5)/2)^2-4))/2`

`color(white)(x) = (-(1+sqrt(5)) +- sqrt((1+sqrt(5))^2-16))/4`

`color(white)(x) = (-(1+sqrt(5)) +- sqrt(2sqrt(5)-10))/4`

`color(white)(x) = -(1+sqrt(5))/4 +- sqrt(10 - 2sqrt(5))/4i`

`color(white)()`
Case `bb(x^2+(1/2-sqrt(5)/2)x+1 = 0)`

Similarly (or simply reversing the signs of the coefficients of `sqrt(5)`)

`x = -(1-sqrt(5))/4 +- sqrt(10 + 2sqrt(5))/4i`

`color(white)()`
Solve `bb(x^5+243) = 0`

Since `243 = 3^5`, the solutions of this are `x = -3alpha` for any `alpha` which is a fifth root of `1`.

So the solutions are:

`x = -3`

`x = (3(1+sqrt(5)))/4 +- (3sqrt(10 - 2sqrt(5)))/4i`

`x = (3(1-sqrt(5)))/4 +- (3sqrt(10 + 2sqrt(5)))/4i`

`color(white)()`
Footnote

Since we were able to construct the `5`th roots of `1` using basic arithmetic operations and square roots, it means that a regular pentagon can be constructed using a straight edge and compass.