Let which satisfy and . Show that .
Any help would be appreciated.
$z_1+z_2+z_3=\exp(\bf{i}\theta_1)+\exp(\bf{i} \theta_2)+\exp(\bf{i}\theta_3)=0$
That does imply that $ \cos(\theta_1)+\cos(\theta_2)+\cos(\theta_3)=0~\&~ \sin(\theta_1)+\sin(\theta_2)+\sin(\theta_3)=0$
because real and imaginary parts are zero.
But what implies anything about $\theta_1+\theta_2+\theta_3~?$
Are you sure that you have given all the details of this question?
Are there other details about the $z_k$?
From the given we do know the set $\{z_1,z_2,z_3\}$ is a subset of the unit circle.
Also we know that $(z_1+z_2+z_3)^2=0$.
Can you rearrange that, see here.
The idea I had in my head which I failed to properly describe is that as these are all unit vectors they are arranged so their vector sum is 0.
Plato properly described this as the two sets of equations relating the cosines and sines of the angles to each other.
Now, given 3 complex numbers we can rotate the unit circle until one of them corresponds with $1+0i$, i.e. $\theta = 0$
Then we can solve for the other two angles given that the cosines and sines sum to zero respectively.
For angles $0 \leq \theta < 2\pi$ there is only one solution
$\theta_1 = 0$
$\theta_2 = \dfrac{2\pi}{3}$
$\theta_3 = \dfrac{4\pi}{3}$
When we square each complex number $\theta$ is doubled and we end up with angles of
$0, ~\dfrac{4\pi}{3},~\dfrac{8\pi}{3}=\dfrac{2\pi}{3}$
and we can see we have our original angles just reordered.
We already know the cosines and sines of these sum to 0 and thus we're done.