I was recently working on the following problem functioning within an overlap of complex numbers and coordinate geometry:
Let $z$ be a complex number $a + ib$ (where $a > b > 0$), and $α_k$ ($0 < k < n + 1$) be the $n^{th}$ roots of unity.
Let $P_k$ be a point on the argand plane such that
$P_k = Re(z).Re(α_k) + i(Im(z).Im(α_k)).$
Prove that the locus of $P_k$ is an ellipse (having $S$ as one of it foci, where $S$ is a point in the argand plane), and
$(i)$ $∑^{n}_{k=1}$ $(P_k - S)$ $= na$
$(ii)$ $∑^{n}_{k=1}$ $(P_k - S)^2$ $= n/2(3a^2 - b^2)$
Here, $P_k$ is set up in terms of $z$, $n$, and $α_k$, but since $z$ and $n$ are both assumed as initial (given) parameters of the question, the locus asked is driven by how $P_k$ moves in the coordinate plane as we vary $k$, up till the given value of $n$.
In my attempt at solving the question, I could not come up with a neater visualization of the problem (which I've noticed is often possible with complex numbers, especially the roots of unity). I tried plugging in the actual points ($z$ as $a + ib$, the general form of the $n^{th}$ roots of unity, etc.) but could not make any headway with that either in determining a locus.
It would greatly help if you could provide a way to visualize and set up the problem, following which I can work through the algebra. Thank you!
For a fixed $z=a+ib$, where we will assume that $a>b>0$.
Let us restate the issue by saying that the immediate generalization of :
$$P_k = \underbrace{\Re(z).\Re(α_k)}_x + i\underbrace{(\Im(z).\Im(α_k))}_y$$
$$\iff P_k =\underbrace{a\Re(α_k)}_x + i\underbrace{(b\Im(α_k))}_y$$
where $\alpha_k=\cos \frac{2 \pi k}{n}+i \sin \frac{2 \pi k}{n}$ is :
$$P_{\theta}=\underbrace{a\cos \theta }_x + i\underbrace{(b \sin \theta)}_y.$$
As $\cos^2 \theta + \sin^2 \theta = 1$,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
which is precisely the equation of an ellipse with : $$d(O,S)=f=\sqrt{a^2-b^2}.$$
Can you take it from here ?
In particular, I think that what you have to prove in (i) is :
$$\sum_{k=1}^n \vec{SP_k} = na \ \text{with} \ S=(-f,0).$$
i.e., to prove that :
$$\sum_k (P_k-S) = \underbrace{\sum_k P_k}_{\color{red}{=0}}-nS = na \ \text{with} \ S=(-f,0),$$
giving therefore $f=a$ but this is not possible because $f<a$. Could you check ?
Dealing with the second sum, when we expand it, we obtain :
$$\sum_k (P_k-S)^2 = \underbrace{\sum_k P_k^2}_{A=0}-2S\underbrace{\sum_k P_k}_{\color{red}{=0}}-nS^2 $$ $$=-nf^2=-n(a^2-b^2)$$
But why do we have $A$ ?
If $n=2m$ is an even number, $A=0$ because it is the sum of the $m$-th roots of unity which are equal to 0.
If $n=2m+1$, as $n \to 2n$ is a bijection modulo $n$ we have still the sum of the $n$th roots of unity which is $0$.