Introduction to De Moivre's theorem

Question

If $a=\cos\frac{2π}7+i\sin\frac{2π}7$, $b=a+a^2+a^4$, $c=a^3+a^5+a^6$, show that $b$ and $c$ are the roots of the equation $x^2+x+2=0$.

Answer 1

HINT

Note that

$$a=\cos\frac{2π}7+i\sin\frac{2π}7=e^{\frac{2πi}7}$$

thus

$$1+a+a^2+a^3+a^4+a^5+a^6=0$$ Proof that sum of complex unit roots is zero

Answer 2

$$b+c=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i+$$ $$+\cos\frac{6\pi}{7}+\cos\frac{10\pi}{7}+\cos\frac{12\pi}{7}+\left(\sin\frac{6\pi}{7}+\sin\frac{10\pi}{7}+\sin\frac{12\pi}{7}\right)i=$$ $$=\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}+$$ $$+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{6\pi}{7}+\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}\right)i=$$ $$=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-1.$$ You can calculate the value of $ab$ by the same way.

Indeed, since we proved that $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2},$$ we obtain: $$bc=\left(-\frac{1}{2}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i\right)\left(-\frac{1}{2}-\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}\right)i\right)=$$ $$=\frac{1}{4}+\left(\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}-\sin\frac{6\pi}{7}\right)^2=$$ $$=\tfrac{1}{4}+\sin^2\frac{2\pi}{7}+\sin^2\frac{4\pi}{7}+\sin^2\frac{6\pi}{7}+2\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}-2\sin\frac{2\pi}{7}\sin\frac{6\pi}{7}-2\sin\frac{4\pi}{7}\sin\frac{6\pi}{7}=$$ $$=\frac{1}{4}+\frac{1}{2}\left(1-\cos\frac{4\pi}{7}+1-\cos\frac{6\pi}{7}+1-\cos\frac{2\pi}{7}\right)+$$ $$+\cos\frac{2\pi}{7}-\cos\frac{6\pi}{7}+\cos\frac{6\pi}{7}-\cos\frac{4\pi}{7}+\cos\frac{4\pi}{7}-\cos\frac{2\pi}{7}=\frac{1}{4}+\frac{7}{4}=2$$ and use the Vieta's theorem.

Also, we can use De Moivre here.

$a^7=\cos2\pi+i\sin2\pi=1$ and since $a\neq1$, we obtain: $$a^6+a^5+a^4+a^3+a^2+a+1=0,$$ which gives $b+c=-1.$

Now, $$bc=a^4(1+a+a^3)(1+a^2+a^3)=a^4(a^6+a^5+a^4+3a^3+a^2+a+1)=2a^7=2$$ and we are done again.