1
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ 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 '16 at 2:09
  • $\begingroup$ @JMoravitz Sorry, fixed. And I know nothing of them. $\endgroup$ – MathMajor Feb 5 '16 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 '16 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 '16 at 2:21
8
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Wow. Can I ask how you spotted that? For future use? $\endgroup$ – MathMajor Feb 5 '16 at 2:17
  • 1
    $\begingroup$ It is a geometric series that he summed. The ratio is $e^z$ $\endgroup$ – Ross Millikan Feb 5 '16 at 2:20
  • 1
    $\begingroup$ Also can use a substitution $u = e^z$. $\endgroup$ – Future Feb 5 '16 at 2:21
  • 2
    $\begingroup$ @MathMajor All polynomials of the form $\sum\limits_{n\ge i\ge 0}x^i$ have that nice property. $\endgroup$ – YoTengoUnLCD Feb 5 '16 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.