Find the roots of z^5+1=0z5+1=0?

2 Answers
Apr 12, 2017

Working to 3dp there are 5 solutions:

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

Explanation:

Let omega=-1 ω=1, and then z^5 = omegaz5=ω

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

|omega| = 1 |ω|=1
arg(omega) = pi arg(ω)=π

So then in polar form we have:

omega = cos(pi) + isin(pi) ω=cos(π)+isin(π)

We now want to solve the equation z^5=omegaz5=ω for zz (to gain 55 solutions):

z^5 = cos(pi) + isin(pi) z5=cos(π)+isin(π)

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 2pi2π, 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

Wolfram AlphaWolfram Alpha

Apr 12, 2017

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.