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$x^4-x^3+x^2-x+1=0$. This equation has to be solved in complex numbers using polar form. I tried grouping the power of x terms but it is not working. Any other idea of solving the equation?

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3 Answers 3

up vote 3 down vote accepted

$$x^4-x^3+x^2-x+1=\frac{x^5+1}{x+1}$$

$$x^5+1=0\implies x^5=-1=e^{i(2n+1)\pi}$$ (using Euler's formula)

where $n$ is any integer and we have $\displaystyle x+1\ne0\iff x\ne-1$

$$\implies\displaystyle x=e^{\frac{i(2n+1)}5}$$ where $n=0,1,3,4$

as $x\ne-1,\displaystyle\iff 2n+1\not\equiv0\pmod5\iff n\not\equiv2\pmod5$

Regarding the general set of values of $n$, see here

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Hints:

$$0=x^4-x^3+x^2-x+1=\frac{x^5+1}{x+1}\;,\;\;x\neq-1$$

and now solve

$$x^5+1=0\iff x^5=-1=e^{\pi i+2k\pi i}\;,\;\;k\in\Bbb Z\;\;\;etc.$$

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Note that \begin{align} x^4-x^3+x^2-x+1=0 &\,\,\,\Longleftrightarrow\,\,\, (x^4-x^3+x^2-x+1)(x+1)=0 \quad\text{and}\quad x\ne -1 \\ &\,\,\,\Longleftrightarrow\,\,\, x^5+1=0 \quad\text{and}\quad x\ne -1. \end{align} Now, solve $x^5+1=0$ using polar form and of course exclude the solution $x=-1$.

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