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I solved the equation $$x^7 - 5=0$$with the help of MATLAB and found seven roots as expected but the thing which wonders me is that all the complex roots have same absolute value aand that is equal to the real root among those seven, what complex roots actually show? Is my hypothesis "All polynomials must have at least one real root" is true? Also calculator gives the value equal to the absolute value of the roots we I solve $\sqrt[7]{5}$ Here is the MATLAB code

x=[1 0 0 0 0 0 0 -5];

b=roots(x)

b =

-1.1339 + 0.5460i

-1.1339 - 0.5460i

-0.2800 + 1.2269i

-0.2800 - 1.2269i

0.7847 + 0.9839i

0.7847 - 0.9839i

1.2585 + 0.0000i

abs(b)

ans =

1.2585

1.2585

1.2585

1.2585

1.2585

1.2585

1.2585

  • 3
    "Is my hypothesis "All polynomials must have at least one real root" is true?" No: $$x^2+1=0$$ – 2017-02-12
  • 3
    If $x^7=5$ then $|x|^7=5$ – 2017-02-12
  • 0
    "What complex roots actually show?" Do you want a plot of those roots ? – 2017-02-12
  • 2
    Your hypothesis need to be modified to be true: All **odd power** polynomials must have at least one real root. – 2017-02-12
  • 1
    @Guangliang: polynomials ... with **real** coefficients – 2017-02-12
  • 1
    Try to put the seven solutions in polar form or exponential form, you may see better of the relationship between the roots. – 2017-02-12

1 Answers 1

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It is well-known that all $n$-th roots of a complex number $z_0$ differ by a factor which an $n$-th root of unity; in other words: once you have a solution $\zeta$ of the equation $z^n=z_0$, all other solutions have the form $$a=\zeta \omega,\quad\text{where}\enspace \omega^n=1.$$