Let $a,b \in \Bbb C$, show that $\sum_{k=0}^{n-1} |a+w^kb|=\sum_{k=0}^{n-1} |b+w^ka|$ with $w=\exp\bigg(\dfrac{2i\pi}{n}\bigg)$

Question

Let $a,b \in \Bbb C$, and let's denote $w=\exp\bigg(\dfrac{2i\pi}{n}\bigg)$

Show that for every $n\ge 2$ $$\sum_{k=0}^{n-1} |a+w^kb|=\sum_{k=0}^{n-1} |b+w^ka|$$ $\bullet~$ My attempts:

I tried to compare the two modules $|a+w^kb|,|b+w^ka|$ for each $k$ and I noticed that $|w|^k=1$ could help me figure it out through some algebraic manipulation but I can't find it. It's easy to show that $$|w^kb-w^ka|=|b-a|$$ But I can't bring each term to the other side because of the module. Any help is greatly appreciated!

Answer 1

Note that $\lvert a+w^kb\rvert=\lvert w^k(b+aw^{n-k})\rvert=\lvert b+aw^{n-k}\rvert$ and so when you sum over $k$ from $0$ to $n-1$, $k$ and $n-k$ hit the same set of values (modulo $n$, which is all that matters for the exponent of an $n$th root of unity).