Question

Find the roots of `z^5+1=0`?

Answer 1

Working to 3dp there are 5 solutions:

`z = -1; -0.309+-0.951; z = 0.809+-0.588i`

Explanation:

Let `omega=-1`, and then `z^5 = omega`

And we will put the complex number into polar form (visually):

`|omega| = 1`
`arg(omega) = pi`

So then in polar form we have:

`omega = cos(pi) + isin(pi)`

We now want to solve the equation `z^5=omega` for `z` (to gain `5` solutions):

`z^5 = cos(pi) + isin(pi)`

Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of `2pi`, so we can equivalently write (incorporating the periodicity):

`z^5 = cos(pi+2npi) + isin(pi+2npi) \ \ \ n in ZZ`

By De Moivre's Theorem we can write this as:

`z = (cos(pi+2npi) + isin(pi+2npi))^(1/5)`
`\ \ = cos((pi+2npi)/5) + isin((pi+2npi)/5)`
`\ \ = cos theta + isin theta \ \ \ \` where `theta=((2n+1)pi)/5`

Working to 3dp we get:

Put:

`n=-2 => theta = -(3pi)/5`
`" " :. z = cos (-(3pi)/5)+ isin (-(3pi)/5)`
`" " :. z = -0.309-0.951i`

`n=-1 => theta = -pi/5`
`" " :. z = cos (-pi/5)+ isin (-pi/5)`
`" " :. z = 0.809-0.588i`

`n=0 => theta = (pi)/5`
`" " :. z = cos (pi/5)+ isin (pi/5)`
`" " :. z = 0.809+0.588i`

`n=1 => theta = (3pi)/5`
`" " :. z = cos ((3pi)/5)+ isin ((3pi)/5)`
`" " :. z = -0.309+0.951`

`n=2 => theta = pi`
`" " :. z = cos pi+ isin pi`
`" " :. z = -1`

After which the pattern continues.

We can plot these solutions on the Argand Diagram

Answer 2

`z_k = e^(i(pi/5+(2kpi)/5)` for `k=0,1,..,4`

Explanation:

If we express `z` in polar form, `z= rho e^(i theta)` we have that:

`z^5 = rho^5 e^(i 5theta)`

so:

`z^5 = -1 => rho^5 e^(i 5theta) = e^(ipi) => {(rho^5 = 1),(5theta =pi+2kpi):}`

We have then:

`{(rho = 1),(theta = pi/5+(2k)/5pi):}`

for any `k in ZZ`

We have potentially infinite solutions:

`z_k=e^(i(pi/5+(2k)/5pi))`

but we can see that if `j-=kmod5`, then `j-k=5n` with `n in ZZ` and we have:

`pi/5+(2jpi)/5 = pi/5+(2(k+5n)pi)/5 = pi/5+(2kpi)/5+2npi`

so that:

`z_j = e^(i(pi/5+(2j)/5pi)) = e^(i(pi/5+(2k)/5pi+2npi)) = e^(i(pi/5+(2k)/5pi))e^(i2npi) = z_k`

In conclusion we have five different solutions for `k=0,1,..,4`:

`z_0 = e^(i(pi)/5) = 1/4((1+sqrt5)+isqrt(10+sqrt5)))`

`z_1 = e^(i(3pi)/5)`

`z_2 = e^(ipi) = -1`

`z_3 = e^(i(7pi)/5)`

`z_4 = e^(i(9pi)/5)`

All this points have module equal to `1`, so they lie on the complex plan on the circle with center the origin and radius `rho = 1`, and they are the vertices of a regular pentagon inscribed in such circle.