How to find real and complex roots of a degree 7?

#x^7=-1#

Find all real and complex roots.

2 Answers
Cem Sentin
Mar 8, 2018

#x_1=Cis(pi/7)#, #x_2=Cis((3pi)/7)#, #x_3=Cis((5pi)/7)#, #x_4=-1#, #x_5=Cis((9pi)/7)#,#x_6=Cis((11pi)/7)# and #x_7=Cis((13pi)/7)#

Explanation:

#x^7=-1#

#x^7=Cis(pi+2pi*k)#

#x=Cis(pi/7+(2pi)/7*k)#

Hence roots of this equation are #x_1=Cis(pi/7)#, #x_2=Cis((3pi)/7)#, #x_3=Cis((5pi)/7)#, #x_4=Cis(pi)=-1#, #x_5=Cis((9pi)/7)#,#x_6=Cis((11pi)/7)# and #x_7=Cis((13pi)/7)#

George C.
Mar 8, 2018

How not to do it...

Explanation:

I would like to show you one way you cannot solve this equation, because it sheds light on the structure of the solution and demonstrates a method that may be useful in other circumstances.

Given:

#x^7 = -1#

We have:

#0 = x^7+1#

#color(white)(0) = (x+1)(x^6-x^5+x^4-x^3+x^2-x+1)#

#color(white)(0) = (x+1)x^3(x^3-x^2+x-1+1/x-1/x^2+1/x^3)#

#color(white)(0) = (x+1)x^3((x+1/x)^3-(x+1/x)^2-2(x+1/x)+1)#

#color(white)(0) = (x+1)x^3(t^3-t^2-2t+1)#

where #t = x+1/x#

So one solution is (unsurprisingly) #x=-1#

The other solutions could be found by solving:

#t^3-t^2-2t+1 = 0#

to find roots #t_1, t_2, t_3#

then solving the three quadratics:

#x^2-t_n x+1 = 0" "# for #n = 1, 2, 3#

which the quadratic formula tells us are:

#x = (t_n+-sqrt(t_n^2-4))/2#

So how do we solve the cubic?

This is where the problem is. The discriminant #Delta# of a cubic of the form #at^3+bt^2+ct+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=-1#, #c=-2# and #d=1#, so we find:

#Delta = 4+32+4-27+36 = 49#

Since #Delta > 0# we can tell that this cubic has three real roots, and since the only possible rational roots #+-1# are not roots, those roots are irrational. This is Cardano's casus irreducibilis. If we attempt to solve this cubic using Cardano's method then we will get a solution expressed in terms of irreducible cube roots of complex numbers.

We could solve the cubic using trigonometric methods, but that does not gain us anything over using trigonometic methods directly with #x^7+1 = 0#. In fact, we would find:

#{ t_1, t_2, t_3 } = { 2cos(pi/7), 2cos((3pi)/7), 2cos((5pi)/7) }#

whereas the roots of #x^7+1=0# are:

#{ -1, cos(pi/7)+-i sin(pi/7), cos((3pi)/7)+-i sin((3pi)/7), cos((5pi)/7)+-i sin((5pi)/7) }#

I say it would not gain us anything, but that's not quite true.

Traditional geometric constructions using straight edge and compasses only allow the construction of regular prime sided polygons if the number of sides is a Fermat prime - i.e. of the form #2^(2^n)+1#. That means that you can (in theory) construct a regular triangle, pentagon, heptadecagon (#17# sides), #257# sided figure and #65537# sided figure. No larger Fermat primes are known. So a regular heptagon is not constructible in this way and neither can we solve the related #x^7+1 = 0#.

However, there are other geometrical constructions using a marked ruler and a technique called neusis, which the Greeks considered an inferior method to straight edge and compass constructions. Also origami allows the solution of cubics and thus the construction of a regular heptagon.

So this reduction of the septic #x^7+1 = 0# to a cubic is actually useful in that it facilitates the solution using any of:

  • Neusis.
  • Origami.
  • Angle trisector.
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