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

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?

• 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? – JMoravitz Feb 5 '16 at 2:09
• @JMoravitz Sorry, fixed. And I know nothing of them. – MathMajor Feb 5 '16 at 2:12
• 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... – JMoravitz Feb 5 '16 at 2:21
• Just substitute $e^z=u$, find that $u$ is a primitive $5$-th root of unity, and take it from there. – Lubin Feb 5 '16 at 2:21

$e^{4z}+e^{3z}+e^{2z}+e^z+1=0$ is too hard to solve directly. Instead, solve $$e^{5z}-1=(e^z-1)(e^{4z}+e^{3z}+e^{2z}+e^z+1)=0$$ and remove roots that are not roots of original equation.
• It is a geometric series that he summed. The ratio is $e^z$ – Ross Millikan Feb 5 '16 at 2:20
• Also can use a substitution $u = e^z$. – Future Feb 5 '16 at 2:21
• @MathMajor All polynomials of the form $\sum\limits_{n\ge i\ge 0}x^i$ have that nice property. – YoTengoUnLCD Feb 5 '16 at 2:41