Question

If $a=e^{i\alpha },b=e^{i\beta },c=e^{i\gamma }$ and $\cos \alpha +\cos \beta +\cos \gamma =0=\sin \alpha +\sin \beta +\sin \gamma$, then prove the following
$a^2+b^2+c^2=0$

Answer 1

$cos\alpha +cos\beta +cos\gamma =0............\left(1\right)sin\alpha +sin\beta +sin\gamma =0..............\left(2\right)Multiplying\left(2\right)\times \left(1\right)andaddingto\left(1\right)(cos\alpha +isin\alpha )+(cos\beta +isin\beta )+(cos\alpha +isiny)=0Let{z}_{1+}{z}_2+z_3=0\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=(cos\alpha -isin\alpha )+(cos\beta -isin\beta )+(cos\gamma -isin\gamma )=0\therefore z_2{z}_3+z_1{z}_3+z_1{z}_2=0{(z_1+z_2+z_3)}^2=z_1^2+z_2^2+z_3^2+2\times 0a^2+b^2+c^2={\left(e^{i\alpha }\right)}^2+{\left(e^{i\beta }\right)}^2+{\left(e^{i\gamma }\right)}^{2}fromdemorestheorem,e^{i\alpha }=cos\alpha +isin\alpha e^{i\beta }=cos\beta +isin\beta and{e}^{i\gamma }=cos\gamma +isin\gamma \therefore a^2+b^2+c^2=0$