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

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#x^7=-1#

Find all real and complex roots.

Find all real and complex roots.

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How not to do it...

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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

So one solution is (unsurprisingly)

The other solutions could be found by solving:

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

to find roots

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

In our example,

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

Since *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

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

whereas the roots of

#{ -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

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 *is* actually useful in that it facilitates the solution using any of:

*Neusis*.- Origami.
- Angle trisector.

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Hence roots of this equation are

Describe your changes (optional) 200