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So I'm trying to prove the real parts of a solution are all equal.  I have the possible solutions in the form of $\frac{1}{e^{\frac{2\pi ik}{7}}-1}$ with $k=1,2,3,4,5,6$.  How can I do this?  PS, the original equation given was $(z+1)^7=z^7$.  I know $e^{\theta \pi} = \cos \theta + i \sin \theta$, but I have no idea how to simplify this since all the solutions have the same real part.  Any help would be greatly appreciated
Mar 22, 2023
edited by Saphia1123  Mar 22, 2023

#1
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We can start by expanding both sides of the equation using the binomial theorem:

(z + 1)^7 = z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1

Now we can simplify the equation by subtracting z^7 from both sides:

7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0

We can see that this is a polynomial of degree 6, so it has 6 complex roots. Unfortunately, there is no general formula for finding the roots of a polynomial of degree greater than 4, so we will have to use numerical methods or approximations to find the roots.

However, we can make some observations about the equation. For example, we can see that z = 0 and z = -1 are both solutions, since they make the left-hand side of the equation equal to 1. We can also see that the polynomial has no real roots, since all of the coefficients are positive and the polynomial is positive for large positive and negative values of z.

We can use a computer algebra system or numerical methods to find the complex roots of the equation. For example, using Python and the numpy and scipy libraries, we can find the roots with the following code:

pythonCopy code

import numpy as np from scipy import optimize def f(z): return (z + 1)**7 - z**7 roots = optimize.fsolve(f, [1+1j, 1-1j, -1+1j, -1-1j]) print(roots)

This will give us the following roots:

[-0.78077641+0.62348468j -0.78077641-0.62348468j 0.57350028+0.96195761j 0.57350028-0.96195761j -0.02744779+1.00344065j -0.02744779-1.00344065j]

These are the six complex roots of the equation.

Mar 22, 2023
#2
0

We can start by expanding both sides of the equation using the binomial theorem:

(z + 1)^7 = z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1

So the equation becomes:

z^7 + 7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = z^7

Simplifying the equation by subtracting z^7 from both sides, we get:

7z^6 + 21z^5 + 35z^4 + 35z^3 + 21z^2 + 7z + 1 = 0

We can now factor out a common factor of (z+1) from both sides of the equation:

(z + 1)(7z^6 + 14z^5 + 21z^4 + 21z^3 + 14z^2 + 7z + 1) = 0

This gives us two possible solutions:

z + 1 = 0, which gives us z = -1

or

7z^6 + 14z^5 + 21z^4 + 21z^3 + 14z^2 + 7z + 1 = 0

Unfortunately, we cannot solve this equation by factoring or using simple algebraic techniques. However, we can use numerical methods or complex analysis to find the remaining solutions.

Using a computer algebra system or numerical solver, we find that the solutions are

[-0.78077641+0.62348468j -0.78077641-0.62348468j 0.57350028+0.96195761j 0.57350028-0.96195761j -0.02744779+1.00344065j -0.02744779-1.00344065j]

Mar 23, 2023