If $\omega^j, j=0,1,2,...,n-1 $ are the $n^{th}$ root of unity, show that $$\sum_{j=0}^{n-1} {|{z_1} + {\omega}^j{z_2}|}^2 = n(|{z_1}|^2 + |{z_2}|^2)$$ for any two complex numbers ${z_1}$ and ${z_2}$.
I did it with two methods. With one I was able to show the result but not with the other. Please tell me what's wrong in my calculation.
Let ${z_1} = x + iy$ and ${z_2} = a + ib$, then $$\sum_{j=0}^{n-1} {|{a + ib} + {\omega}^j{(x + iy)}|}^2$$ $$ = \sum_{j=0}^{n-1} ({a + {\omega}^jx})^2 + (b+{\omega}^jy)^2$$ $$ = \sum_{j=0}^{n-1} a^2 + b^2 + \omega^{2j}x^2 + \omega^{2j}y^2 + 2ax\omega^j + 2ay\omega^j$$ $$ = \sum_{j=0}^{n-1} a^2 + b^2 + \sum_{j=0}^{n-1} (x^2+y^2)\omega^{2j} + (2ax + 2ay)\sum_{j=0}^{n-1}\omega^j$$
last summation will be zero because it's the sum of roots of unity and so we get, $$ n(a^2 + b^2) + \sum_{j=0}^{n-1} (x^2+y^2)\omega^{2j}$$
Now I don't know what to do, like if last summation is also zero then what is wrong in my calculation because answer won't come if it is zero and if it is not then what is the value of last summation and how to get it.
$$\sum_{j=0}^{n-1}|{a + ib} + {\omega}^j{(x + iy)}|^2 =\sum_{j=0}^{n-1} ({a + {\omega}^jx})^2 + (b+{\omega}^jy)^2$$ is wrong because the $\omega^j$ are not real numbers, so $$ (a + \omega^j x) , (b + \omega^j y) $$ are not the real and imaginary part of ${a + ib} + {\omega}^j{(x + iy)}$.
For a correct derivation you can use that $$ | z + w| ^2 = |z|^2 + |w|^2 + 2 \operatorname{Re}(\bar z w) $$ for complex numbers $z, w$: $$ \sum_{j=0}^{n-1} |z_1 + \omega^j z_2|^2 = \sum_{j=0}^{n-1} \bigl(|z_1|^2 + |\omega^j z_2|^2 + 2 \operatorname{Re}(\overline z_1 \omega^j z_2) \bigr)\\ = n |z_1|^2 + n|z_2|^2 + 2 \operatorname{Re}\left(\sum_{j=0}^{n-1} \overline z_1 \omega^j z_2\right) \\ = n |z_1|^2 + n|z_2|^2 + 2 \operatorname{Re}\left(\overline z_1 z_2\sum_{j=0}^{n-1} \omega^j\right) $$ and $ \sum_{j=0}^{n-1} \omega^j = 0$.