Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities?

$$z^5=1$$ $$z^5-1=0$$ $$(z-1)(z^4+z^3+z^2+z+1)=0$$ Note that the parenthesis with 5 terms are the roots but as we know the complex is a filed no zero divisors thus the sum of the roots is zero! Plug in $e^{frac{2pi i }{5}}=z$ $$(e^{frac{2pi i }{5}}-1)(e^{frac{8pi i }{5}}+ e^{frac{6pi i }{5}}+ e^{frac{4pi i }{5}}+ e^{frac{2pi i }{5}}+1)=0$$

Let $S$ be the set of roots of $z^5-1$, then $$e^{frac{2pi i}{5}}S={e^{frac{2pi i}{5}}, e^{frac{4pi i}{5}},e^{frac{6pi i}{5}},e^{frac{8pi i}{5}},e^{frac{10pi i}{5}}}={e^{frac{2pi i}{5}},e^{frac{4pi i}{5}},e^{frac{6pi i}{5}},e^{frac{8pi i}{5}},1}=S$$

The sets are equal, so $$e^{frac{2pi i}{5}}sum_{zin S}z=sum_{zin S}z\Rightarrow sum_{zin S}z=0$$

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More generally

Let $G$ be a multiplicative group of order $n$. Then $$sum_{zin G}z=0$$

Proof: Let $gin G$ such that $gneq e$. Let $phi_g(x)=gx$, then $phi_gintext{Aut}(G)$ since:

• injective: If $phi_g(x)=phi_g(y)$ then $gx=gyRightarrow x=y$.
• surjectivity: If $ain G$, then $phi_g(g^{-1}a)=a$.

This means that the sets $gG,G$ are equal. And so,

$$sum_{zin G}gz=sum_{zin G}giff (g-e)sum_{zin G}z=0\Rightarrow sum_{zin G}z=0$$

Good question. Shall we try?

So we have the five roots: $$1,left(cos left(frac{2 pi}{5}right)+i sin left( frac{2pi}{5}right)right),left(cos left(frac{4 pi}{5}right)+i sin left(frac{4 pi}{5}right)right),left(cos left(frac{6 pi}{5}right)+i sin left(frac{6 pi}{5}right)right), left(cos left(frac{8 pi}{3}right)+i sin left(frac{8 pi}{5}right)right)$$

I am just going to write $frac{2pi}{5}$ as $alpha$. It's going to save time and make this easier to read. In this notation, the roots are : $1, cos (alpha) + icdotsin(alpha), cos (2alpha) + icdotsin(2alpha), cos (3alpha) + icdotsin(3alpha)$ and $cos (4alpha) + icdotsin(4alpha)$

OK let's add them all up, we'll start with the imaginary part because that is going to be easy. We have $$i times left[sin alpha+sin 2alpha+sin 3alpha+sin 4alpharight]$$ Well we know that $5alpha = 2pi$ and so $sin alpha = sin(2pi-alpha) = sin(-4 alpha) =-sin (4alpha$). Also $sin(2 alpha)=-sin(3 alpha)$ for similar reasons and so all of these terms cancel to zero.

In the real part, we have $$1+cos(alpha)+cos(2alpha)+cos(3alpha)+cos(4alpha)$$ $cos(alpha)=cos(2pi-alpha)=cos(4alpha)$ Also $cos(2 alpha)=cos(3 alpha)$ for similar reasons. So we just need to show that $$1+2cos(alpha)+2cos(2alpha)=0$$

How hard can this be? We know, from the trig identities that $cos(2alpha)=2 cos^{2} (alpha)-1$. So now we need $1+2 cos(alpha)+4 cos^{2}(alpha)-2=0$ or $$4 cos^{2}(alpha)+2 cos(alpha)-1=0$$

Trigonometry tells us that $cos alpha = frac{sqrt 5-1}{4}$. If we pop this in then we get.

$$frac{6-2 sqrt 5}{4}+frac{sqrt 5-1}{2}-1=0$$

Hooray and Phew.

If you want a simple trigonometric proof, recall $$sin a-sin b=2sinfrac{a-b}2cosfrac{a+b}2.$$ Therefore begin{align} 0&=sin(a+2pi)-sin a\ &=(sin(a+2pi)-sin(a+8/pi/5))+(sin(a+8/pi/5)-sin(a+6/pi/5))\ &+(sin(a+6/pi/5)-sin(a+4/pi/5))+(sin(a+4/pi/5)-sin(a+2/pi/5))\ &+(sin(a+2/pi/5)-sin a)\ &=2(sinpi/5)left[cos(a+9pi/5)+cos(a+7pi/5)+cos(a+pi)+cos(a+3pi/5)+cos(a+pi/5)right] end{align} and $$cos(a+9pi/5)+cos(a+7pi/5)+cos(a+pi)+cos(a+3pi/5)+cos(a+pi/5)=0.$$ Taking $a=-pi/5$ gives $$cos(8pi/5)+cos(6pi/5)+cos(4pi/5)+cos(2pi/5)+1=0.$$ and taking $a=-7pi/10$ gives $$sin(8pi/5)+sin(6pi/5)+sin(4pi/5)+sin(2pi/5)+0=0.$$ These are the real and imaginary parts of the sum of the fifth roots of zero.

Of course, this method works for other values of $5$.

You don't need "geometric sum" to prove $$ sum_{j=0}^{n-1} cos(theta+jphi)=frac{sum_{j=0}^{n-1}cos(theta+jphi)sin(phi/2)}{sin(phi/2)}=frac{sum_{j=0}^{n-1}left[sin(theta+(j+tfrac12)phi)-sin(theta+(j-tfrac12)phi)right]}{sin(phi/2)}=dots $$ and similarly $sum_{j=0}^{n-1}sin(theta+jphi)$.