240 Followers Problem - 3

Let $a,b,c$ be three complex numbers such that $a=e^{\theta_1 i}$ , $b=e^{\theta_2 i}$ and $c=e^{\theta_3 i}$ . Then let's find the magnitude of their sum, i.e., $|a+b+c|$ :

$|a+b+c|=|\cos \theta_1+\cos \theta_2+\cos \theta_3+i(\sin \theta_1+\sin \theta_2+\sin \theta_3) \\ |a+b+c|=\sqrt{(\cos \theta_1+\cos \theta_2+\cos \theta_3)^2+(\sin \theta_1+\sin \theta_2+\sin \theta_3)^2} \\ |a+b+c|=\sqrt{3+2(\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2+\cos \theta_2 \cos \theta_3+\sin \theta_2 \sin \theta_3+\cos \theta_3 \cos \theta_1+\sin \theta_3 \sin \theta_1)} \\ |a+b+c|=\sqrt{3+2(\cos(\theta_1-\theta_2)+\cos(\theta_2-\theta_3)+\cos(\theta_3-\theta_1))} \\ |a+b+c|=\sqrt{3+2\left(-\dfrac{3}{2}\right)} \\ |a+b+c|=0 \implies a+b+c=0$

Hence, $\cos \theta_1+\cos \theta_2+\cos \theta_3+i(\sin \theta_1+\sin \theta_2+\sin \theta_3)=0$ . Comparing the imaginary part we get $\boxed{0}$ as our answer.

$The ~ sum ~ given~ = - 1\frac 1 2,~~ |Cos\theta|~is~never~greater~than~1,~~ the~ sum~~ is~~ negative,\\we~also~have~~ |Cos(60, ~120~240,OR~~300)|=\dfrac 1 2.\\ \therefore~ - 1\frac 1 2=- \frac 1 2 - \frac 1 2 - \frac 1 2, ~~~~OR~~~~ 0, - \frac 1 2, - 1~~OR~~\dfrac12,~- 1, - 1.\\ \text{As per Niven's theorem, there is no other angle that would give rational number, and }1\frac 1 2, is ~rational.\\ Let~us~try~the~1^{st}.~~\therefore \text{ the differences must be -120 and -240.}\\ \text{Let us take one angle as 60. The other two must be } 60\pm 120,~~~ or~~~60\pm 240. \\ \therefore~\theta_1=60^o, ~~\theta_2=180^o, ~~\theta_3=300^o,\\ !!!! ~It~works!\\ Sin60+Sin180+Sin300=\Large ~~~~\color{#D61F06}{0}$

Given $\cos \left( \theta_{1} -\theta_{2} \right) + \cos \left( \theta_{2} -\theta_{3} \right) + \cos \left( \theta_{3} -\theta_{1} \right) = -\dfrac{3}{2}$

$sin(\theta_{1})sin(\theta_{2})+cos(\theta_{1})cos(\theta_{2})+sin(\theta_{2})sin(\theta_{3})+cos(\theta_{2})cos(\theta_{3})+sin(\theta_{3})sin(\theta_{1})+cos(\theta_{3})cos(\theta_{1})=-\frac{3}{2}$

$sin(\theta_{1})+sin(\theta_{2})+sin(\theta_{3})=\sqrt{3+2(sin(\theta_{1})sin(\theta_{2})+cos(\theta_{1})cos(\theta_{2})+sin(\theta_{2})sin(\theta_{3})+cos(\theta_{2})cos(\theta_{3})+sin(\theta_{3})sin(\theta_{1})+cos(\theta_{3})cos(\theta_{1}))-(cos(\theta_{1})+cos(\theta_{2})+cos(\theta_{2}))^{2})}$

$\sqrt{-(cos(\theta_{1})+cos(\theta_{2})+cos(\theta_{3}))^{2}}$

Therefore only possible answer is 0.