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Let P be a point that lies along the real axis (i.e., x axis) a distance d from the origin, where d < 1. Using the results in part (a) $\sum_{k=0}^{n-1} = e^{j*2\pi/n*k} =0$ that the product of the distances from the point P to each of the n vertices of a regular n-gon of radius 1 centered at the origin with one point on the real axis is given by $1−d^n$.

I think I set it up correctly $$\prod_{k = 0}^{n-1} |e^{j2\pi/nk} - d|$$

I tried Euler's formula, and I got it into this form: $$\prod_{k = 0}^{n-1} \sqrt{1-2d\cos(2\pi/nk)+d^2}$$ and I don't know how to go from here. Any suggestions? I also don't know how to use the fact that the sums of the sides of the polygon add to 0.

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    `the regular n-gon of radius 1` There are many regular $n$-gons with radius $1$ in the complex plane. Which one does the question ask about? – 2017-01-09
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    @dxiv All of them. Assume an arbitrary number of sides: n – 2017-01-09
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    The result you propose is not true if, for example, the $n$-gon is a square centered at $10+10i$ and rotated $36^\circ$ degrees counterclockwise. Again, which particular regular $n$-gon is the question about? – 2017-01-09
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    @dxiv Oh, yes I forgot one of the conditions. the original n-gon must have one side on the real axis. Thanks for pointing it out. – 2017-01-09
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    I think it is $2\pi k/n$, not $2\pi/(nk)$. – 2017-01-09
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    @Goldname The question is still not complete unless you add that the polygon is centered at the origin (or, equivalently, inscribed in the unit circle). – 2017-01-10
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    @dxiv corrected – 2017-01-10

1 Answers 1

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Hint: given the radius of $1\,$, the vertex of the regular $n$-gon which lies on the real axis can only be at $\pm 1$. Assume it is at $+1$, then the $n$ vertices are the $n^{th}$ roots of unity satisfying $z^n-1=0$. Let the roots be $\{\omega_k \;|\; k = 0 \dots n-1\}\,$ then $P(z)=z^n-1=\prod_{k-0}^{n-1}(z-\omega_k)$.

It follows that $\prod_{k-0}^{n-1}|d-\omega_k|=\left|\prod_{k-0}^{n-1}(d-\omega_k)\right| = \left|P(d)\right| = \left|d^n-1\right| = 1 - d^n$ where the latter equality comes from $|d| \le 1\,$.

Similar argument proves the same result if the vertex is at $-1$ instead (or, it can be shown to follow directly by symmetry).

[ EDIT ]  To clarify the steps, as asked in a comment:

  • The relation $z^n-1=\prod_{k-0}^{n-1}(z-\omega_k)$ holds true for all $\forall z \in \mathbb{C}\,$. The LHS is essentially the definition of the $n^{th}$ roots of unity, while the RHS follows directly from the unique factorization of polynomials in $\mathbb{C}[\text{X}]\,$.

  • Taking the absolute value on both sides: $|z^n-1|=\left|\prod_{k-0}^{n-1}(z-\omega_k)\right|=\prod_{k-0}^{n-1}|z-\omega_k|\,$. This again holds true for all $\forall z \in \mathbb{C}\,$, and means that the product of distances from any point $z$ in the complex place to the vertices of the regular polygon in question equals $|z^n-1|$.

  • Specializing the above for the case $\,z=d\,$ with $\,d \in \mathbb{R}, |d| \lt 1\,$gives $\prod_{k-0}^{n-1}|z-\omega_k|=|d^n-1|$.

  • Finally, since $|d| \lt 1$ it follows that $|d|^n \lt 1 \implies d^n \le |d|^n \lt 1\,$ so $|d^n-1| = 1 - d^n$.

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    Thanks for the response, but I am not really understanding where $z^n-1$ = 0$ came from? – 2017-01-09
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    @Goldname An $n^{th}$ root of unity is by definition a root of $z^n=1 \iff z^n-1=0$. Also, by DeMoivre's formula, the $n^{th}$ roots of unity are the vertices of a regular $n$-gon inscribed in the unit circle, with one vertex at 1 (see [here](https://en.wikipedia.org/wiki/Root_of_unity#Trigonometric_expression) for example). – 2017-01-09
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    Sorry for the late response. I looked at your answer some more. I don't understand how you know that d is a root of unity. How can $|\Pi(d-k)| = |P(d)|$ when d is an arbitrary number, not a root of unity? – 2017-01-10
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    @Goldname $d$ is *not* a root of unity, and the proof nowhere assumes that. But $\prod(d-\omega_k)=P(d)$ because of the equality established at the first step, and then taking the absolute value of both sides completes the proof. I edited my answer to break down the steps and (hopefully) clarify it. – 2017-01-10
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    The third bullet point is where I am not understanding. You can't assign z = d, because z is defined as a root of unity. d is not a root of unity, right? So you cannot generalize z because z is restricted to be a root of unity. – 2017-01-10
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    @Goldname `z is defined as a root of unity` No it's not, where did you get that idea? The roots of unity are $\omega_k$. $z$ is the free variable in the polynomial $P \in \mathbb{C}[\text{X}]$. To evaluate the polynomial for a particular complex value, you assign $z$ the respective value, just like you do with all polynomials. In this case we are interested in the value of $P(z)$ at $d$ so we assign $z=d\,$. – 2017-01-10
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    Perhaps I am misunderstanding. I thought that z is a root of unity if it satisfied this $z^n =1$, right? – 2017-01-10
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    @Goldname $z^n-1$ is the polynomial, $z^n=1$ is the equation which gives its roots, and $\omega_k$ are those roots (which in the case of this polynomial are called the $n^{th}$ roots of unity). $z$ itself is *not* a root of unity. Simple case $n=2\,$: the polynomial is $z^2 -1$, and the equation $z^2=1$ gives the roots $\pm 1$ ***but*** the polynomial identity $z^2-1=(z-1)(z+1)$ holds for *all* $z$ and can be evaluated for *all* $z$ e.g. at $z=3\,$: $3^2-1=(3-1)(3+1)$. – 2017-01-10
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    Yes I understand now. But just to confirm I am assuming that any complex number with radius = 1 is a root of unity? – 2017-01-10
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    @Goldname No, the number of $n^{th}$ roots of unity is exactly for any given $n$. For example the $4^{th}$ roots of unity are $\{\pm1, \pm i\}$ and the $2^{nd}$ (square) roots of unity are $\{\pm1\}$ so $i$ is a $4^{th}$ root of unity, but it is not a square root of unity. What's true is the converse: any root of unity of any order has absolute value $1$. – 2017-01-10
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    Then what if one of the vertices of the polygon is not a root of unity? The questions says the points have a radius of one. The question never says that the points are a root of unity. – 2017-01-10
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    @Goldname The question says it's a regular polygon centered at the origin with one vertex at $\pm 1$ (and from the question, it sounded like you were familiar with De Moivre's formula). Refer for example to [Root of unity - Trigonometric expression](https://en.wikipedia.org/wiki/Root_of_unity#Trigonometric_expression): "*De Moivre's formula ... shows that on the complex plane the nth roots of unity are at the vertices of a regular n-sided polygon inscribed in the unit circle, with one vertex at 1*". – 2017-01-10