Find the roots of z5+1=0?

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Aiden Cooper · Accepted Answer

Step 1  Let ω=−1, and then z5=ω  And we will put the complex number into polar form (visually):  |ω|=1  arg(ω)=π  So then in polar form we have:  ω=cos⁡(π)+isin⁡(π)  We now want to solve the equation z5=ω for z (to gain 5 solutions):  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 2π, so we can equivalently write (incorporating the periodicity):  z5=cos⁡(π+2nπ)+isin⁡(π+2nπ)n∈Z  By De Moivres

Karli Kaiser · Accepted Answer

Step 1  If we express z in polar form, z=ρeiθ we that:  z5=ρ5ei5θ  so:  z5=−1⇒ρ5ei5θ=eiπ  ⇒{ρ5=15θ=π+2kπ  We have then:  {ρ=1θ=π5+2k5π  for any k∈Z  We have potentially infinite solutions:  zk=ei(π5+2k5π)  but we can see that if j≡k mod 5, then j−k=5n with n∈Z and we have:  π5+2jπ5=π5+2(k+5n)π5=π5+2kπ5+2nπ  so that:  zj=ei(π5+2j5π)=ei(π5+2k5π+2nπ)=ei(π5+2k5π)ei2nπ=zk  In conclusion we have five different solutions for k=0,1,⋯,4  z0=eiπ5=14((1+5)+i10+5)  z1=ei3π5  z2=eiπ=−1  z3=ei7π5  z4=ei9π5  All this points have module equal to 1, so they lie on the complex plan on the circle with center the origin and radius ρ=1, and they are the vertices of a regular pentagon inscribed in such circle.

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

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