roots of unity and geometry problem

Question

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I have managed to do part A but part B's starting point is confusing. If the question stated find the sum of the distance from X to the vertices of the polygon the rest of the question is fine.

But why does mean the sum of XA mean finding the sum of|w-2|

Answer 1

The first version of this solution was incorrect on its final steps. An edit was made based on user @dxiv's comments below.

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From the first relationship, $|w-2|^2=5-2(w+\bar w)$, if w is a root of unity, we have: $$\sum_{i=1}^7(XA_i)^2=\sum_{i=1}^7[5-2(A_i-\bar A_i)]$$ since every $A_i$ is a root of unity.

So $$S=\sum_{i=1}^7[5-2(A_i-\bar A_i)]=35-4i\Im\Big(\sum_{i=1}^7 A_i\Big)$$

since $z-\bar z=2i\Im(z)$.

But as @dxiv observed in the comments, since the LHS is a real number, we must have $$\Im\Big(\sum_{i=1}^7 A_i\Big)=0$$

Thus, $S=35$.

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Answer 2

Let $z\in\mathbb C$ and $|z|=1$. See that $$|z-2|^2 = (z-2)(\overline{z-2})=(z-2)(\overline{z}-\overline{2}) \\ =z\overline{z}-2z-2\overline{z}+4 \\ =|z|^2-2(z+\overline{z})+4 \\ =1+4-4\left(\frac{z+\overline{z}}{2}\right) \\ =5-4\Re (z)$$ Which is obviously equal to the Right Hand Side.