Let $z_k = \cos\frac{2k\pi}n + i\sin\frac{2k\pi}n$. Show that $\sum_{k=1}^n|z_k-z_{k-1}|<2\pi$. 
Let $z_k = \cos\frac{2k\pi}n + i\sin\frac{2k\pi}n$. Show that $\sum_{k=1}^n|z_k-z_{k-1}|<2\pi$.

So, this is a problem from Herb Silverman's "Complex Variables".
I gave it a try with several ways, e.g., reducing all terms to one term, but none of them work. It seems that I never got close.
Can someone help me?
 A: For $k = 6$, you should think of the problem in terms of the following picture:
$\hspace{44mm}$
Thinking of the circle as the unit circle, the vertices of the hexagon are the points $z_0, \dots, z_5$ and the lengths of the sides of the hexagon are given by $|z_k - z_{k-1}|$, so the perimeter is given by the sum $\sum_{k=0}^5|z_k - z_{k-1}|$. The circumference of the circle which inscribes the pentagon is $2\pi$.
For other values of $n$, you obtain a similar picture with the hexagon replaced by an $n$-gon.
One way to establish the inequality $\sum_{k=0}^{n-1}|z_k - z_{k-1}| < 2\pi$ is to prove that $|z_k - z_{k-1}| < \frac{2\pi}{n}$. In order to do this, think of two paths between $z_{k-1}$ and $z_k$, then calculate their lengths. Which one is necessarily shorter?
A: I'm going to begin by making an identification of the $z_k$ numbers which makes the inequality pretty intuitive before we show it rigorously.
The set $\{z_k\}$ for $n$ are, you might recognize, the so called roots nth roots of unity. These are the solutions of the equation $z^n = 1$ and as you might remember from solving such equations, or seeing visual representationon wolfram alpha http://www.wolframalpha.com/input/?i=z%5E7+%3D+1,  they lie on the unit circle in a polygonal pattern. $|z_{k} - z_{k-1}|$ is the distance between adjacent number on the circle, that is the length of the sides of the polygon and the sum becomes the total circumference of the polygon.
The fact that the circumference of a polygon inscribed in a circle is less than the circumference of the circle, in this case $2\pi$, is pretty intuitive. 
To show it you need, in this case, only show
$$|z_k - z_{k - 1}| < \frac{2\pi}{n},$$ that is that the chord is shorter than the than the corresponding arc.
Geometrically this follows from the notion that the line is the shortest path between two points but if we want a purely algebraic(calculus) proof we may do the following:
If we use exponential notation $z_k = e^{2\pi i  k/n}$
$$|z_k - z_{k - 1}| = |e^{2\pi i  k/n} - e^{2\pi i  (k - 1)/n}| = |e^{2\pi (k - 1)/n}(e^{2\pi i/n} - 1)|  = |e^{2\pi i/n} - 1|$$
$$|e^{2\pi i/n} - 1|^2 = (e^{2\pi i/n} - 1)(e^{-2\pi i/n} - 1) = 2(1 - \cos(2\pi/n))$$
In the next step we use the cosine inequaity which may be stated as $1 - \cos x < x^2/2$ or $2(1 - \cos x) \leq x^2$
$$|z_k - z_{k - 1}| = |e^{2\pi i/n} - 1| = \sqrt{2(1 - \cos (2\pi / n))}\leq \sqrt{(2\pi / n)^2 } = \frac{2\pi}{n}$$
which is precisely what we expected.
An neat generalization of this problem would be to drop the specific form of the numbers entirely and just say that $(z_k)$ is a finite sequence of complex numbers on the unit circle which are numbered in the order they appear along the circle right to left. The inequality 
$$\sum_{k = 1}^n |z_{k} - z_{k-1}|\leq 2\pi$$
should still hold by the "circumference of polygon is less than circumference of circle" argument but the algebraic proof would need to be modified. (You'd make the identification $z_0 = z_n$ to close the loop)
