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Solve $$e^{4z} +e^{3z} + e^{2z} + e^z + 1 = 0.$$

I have attempted this problem with the usual definition by writing $z=x+iy$ and using $e^z = e^x(\cos y + i \sin y)$ but got a large unsolvable mess. Is there something I'm missing?

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    $\begingroup$ You have an $x$ in the exponent of the first term while $z$'s in the exponent of the remaining terms. Is this intentional? What do you know about roots of unity and their sums? $\endgroup$
    – JMoravitz
    Feb 5, 2016 at 2:09
  • $\begingroup$ @JMoravitz Sorry, fixed. And I know nothing of them. $\endgroup$
    – MathMajor
    Feb 5, 2016 at 2:12
  • $\begingroup$ To elaborate on my hint, the $n^{th}$ roots of unity are the $n$ solutions to the equation $z^n=1$, or written another way, $z^n-1=0$. One learns, from efforts like these that the sums of the roots of unity is zero. If $e^z$ happens to be a primitive fifth root of unity (i.e. not equal to one), then... $\endgroup$
    – JMoravitz
    Feb 5, 2016 at 2:21
  • $\begingroup$ Just substitute $e^z=u$, find that $u$ is a primitive $5$-th root of unity, and take it from there. $\endgroup$
    – Lubin
    Feb 5, 2016 at 2:21

1 Answer 1

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$e^{4z}+e^{3z}+e^{2z}+e^z+1=0$ is too hard to solve directly. Instead, solve \begin{equation} e^{5z}-1=(e^z-1)(e^{4z}+e^{3z}+e^{2z}+e^z+1)=0 \end{equation} and remove roots that are not roots of original equation.

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  • $\begingroup$ Wow. Can I ask how you spotted that? For future use? $\endgroup$
    – MathMajor
    Feb 5, 2016 at 2:17
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    $\begingroup$ It is a geometric series that he summed. The ratio is $e^z$ $\endgroup$ Feb 5, 2016 at 2:20
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    $\begingroup$ Also can use a substitution $u = e^z$. $\endgroup$
    – Future
    Feb 5, 2016 at 2:21
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    $\begingroup$ @MathMajor All polynomials of the form $\sum\limits_{n\ge i\ge 0}x^i$ have that nice property. $\endgroup$ Feb 5, 2016 at 2:41

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