Roots of Unity: second largest value and absolute value

Question

Consider the $n$th roots of unity $e^{2 \pi i k/n}$ for fixed integer $n \geq 2$ and $0 \leq k < n$.

Now I am interested in the second largest value (in absolute value) of the values

$$\frac{e^{2 \pi i k/n} + 1} {2}, \quad 0 \leq k < n.$$

For which $k$ does this happen and what is the absolute value?

In my lecture notes it is written that this value is given for $k=1$ with absolute value

$$\sqrt{\frac{1}{2} + cos(2 \pi /n) /2}.$$

Can you help me how to get to this result. I also wonder whether this is true for $n$ even and $n$ odd.

Answer 1

You can answer this question geometrically. Since division by 2 does not affect the order of the norms, we can find the second largest value of $$|| e^{2\pi i k/n} + 1 ||.$$ For $0 \leq k < n$, $e^{2\pi i k/n}$ lies on the unit circle, and adding by 1 shifts the circle 1 unit to the right. Therefore, the largest norm would be at $k = 0$ (the $0^\circ$ angle value), and the second largest would be at $k = 1$ and $k = n-1$ (by symmetry). Taking $k = 1$, we have $$\frac{|| e^{2\pi i/n} + 1 ||}{2} = \frac{\sqrt{(\cos\left(\frac{2\pi}{n}\right)+1)^2 + \sin^2\left(\frac{2\pi}{n}\right)}}{2}.$$ Simplifying gives your answer. A lot of problems in complex variables can be solved with geometric intuition.

Answer 2

Since all roots of unity have the same absolute value, this is the same as asking which is the second closest root of unity to $1$ (note that the closest one is always $1$ itself).

Note that you can always represent the $n$-th roots of unity as equally spaced points on the complex unit circle, including $1$. The geometric picture makes it clear that there are exactly two solutions to your problem: $$ e^{2i\pi/n} \qquad \text{and} \qquad e^{2i\pi(n-1)/n} $$

Answer 3

First of all let us note that the largest absolute value is obtained for $k=0$, which gives the real number $1$, as it has just been pointed out by @AlexR. The second largest absolute value is obtained for the angle that is closest to $0$ (or $2 \pi$), which is $k=1$. Note that you are adding two complex numbers, and one of them is $1$, which is real. The smallest the angle between the numbers, the largest the absolute value of their sum.

Answer 4

Let $z=e^{2\pi i k/n}$. Then $$ \begin{aligned} |1+z|^2&=(1+z)(1+\overline{z})=1+(z+\overline{z})+z\overline{z}\\ &=2+2\Re z\\ &=2+2\cos(2\pi k/n). \end{aligned} $$ Because cosine is a decreasing function in the interval $[0,\pi]$ it is clear that $k=0$ gives the largest value, $k=1$ the next to largest et cetera (up to $k\le n/2$). Because cosine is also an even function with period $2\pi$ this is the only range of values of $k$ you need to consider.

Answer 5

$$\left|\frac{e^{2 \pi i k/n} + 1} {2}\right| = \frac{\left|e^{2 \pi i k/n} + 1\right|} {2} = \frac{\left|\cos(2 \pi k/n) + i\sin(2 \pi k/n) + 1\right|} {2} =\frac{\left|\cos(2 \pi k/n) + 1 +i\sin(2 \pi k/n)\right|} {2} = \frac{1}{2} \sqrt{(\cos(2 \pi k/n) + 1)^2 + \sin^2(2 \pi k/n)} = \frac{1}{2} \sqrt{(2+2\cos(2 \pi k/n)} = \sqrt{\frac12+\frac12\cos(2 \pi k/n)}$$

This is maximized where $\cos(2 \pi k/n)=1$, or wherever $k/n \in \mathbb{Z}$. Letting $k=0$ would be the simplest way to achieve this. Since we want the next largest value, we do not wish to alter $k$ by much at all. The smallest deviation is by $\pm1$.