Animated Solution for Mathematics - Complex Numbers: Prove that ${\sum }_{k=1}^{n-1}(n-k)\cos \frac{2k\pi }{n}=-n/2$, where $n\ge 3$ is an integer.

A Journey into Complex Sums

Welcome, student. Today, we are not just solving a summation; we are uncovering a hidden symmetry in the unit circle. When you look at the expression ${\sum }_{k=1}^{n-1}(n-k)\cos \frac{2k\pi }{n}$, you might see a daunting trigonometric series.

But I want you to see something else. I want you to see vectors. I want you to see the roots of unity dancing around the origin.

The Complex Transformation

Whenever you see a trigonometric sum where the angles are in an arithmetic progression, your intuition should immediately scream: "Complex Numbers!" Because trigonometry is just the projection of complex rotation.

Let us define $z=e^{i\frac{2\pi }{n}}$. This $z$ is the primitive $n$-th root of unity. By Euler's formula, $z=\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$.

Our target sum is the real part of a complex sum $S$. Let us define:

Notice that $\text{Re}(S)={\sum }_{k=1}^{n-1}(n-k)\text{Re}(z^k)={\sum }_{k=1}^{n-1}(n-k)\cos \frac{2k\pi }{n}$. We have successfully translated our problem from the messy world of cosines into the clean, algebraic world of complex powers.

The Decomposition

Now, let us expand $S$. We can distribute the $(n-k)$ term:

Let us call these two parts $S_1$ and $S_2$. For $S_1$, we recall the beautiful property of the roots of unity: the sum of all $n$ roots is zero, i.e., ${\sum }_{k=0}^{n-1}{z}^k=0$.

Since our sum $S_1$ starts from $k=1$, we are missing the $k=0$ term, which is $z^0=1$. Thus, ${\sum }_{k=1}^{n-1}{z}^k=-1$. This simplifies $S$ to $S=-n-S_2$.

The AGP Battle

Now, we face $S_2={\sum }_{k=1}^{n-1}k{z}^k$. This is an Arithmetico-Geometric Progression. To solve it, we use the "shift and subtract" method.

We write $S_2=z+2z^2+3z^3+\cdots +(n-1)z^{n-1}$. Then, we multiply by $z$:

Subtracting these two equations:

Since $z^n=1$, the last term is simply $-(n-1)$. The geometric series $z+z^2+\cdots +z^{n-1}$ is again $-1$. Thus:

This gives us $S_2=\frac{-n}{1-z}=\frac{n}{z-1}$.

The Half-Angle Magic

We are at the final stretch. We have $S=-n-S_2=-n-\frac{n}{z-1}$. Simplifying this, we get:

To extract the real part, we use the half-angle trick. Let $\theta =\frac{2\pi }{n}$. Then:

Multiplying numerator and denominator by $e^{-i\theta /2}$, we get:

Finally, $S=-n\left(\frac{1}{2}-\frac{i}{2}\cot (\frac{\pi }{n})\right)$. The real part is clearly $-\frac{n}{2}$.

We have arrived at the destination. The complexity collapses into a simple constant. This is the beauty of JEE mathematics—the most complex problems often yield the most elegant, simple truths.