# Geometry in complex numbers.

Let $$\theta_1, \theta_2, \theta_3, … , \theta_{10}$$ be positive valued angles (in radian) such that $$\theta_1+\theta_2+ \theta_3+… + \theta_{10}=2\pi$$. Define complex numbers $$z_1=e^{i\theta_1}$$, $$z_k=z_{k-1}e^{i\theta_k}$$ for $$2\leq k\leq 10$$. Then which of these is true?
P: $$|z_2-z_1|+|z_3-z_2|+…|z_1-z_{10}|\leq 2\pi.$$
Q: $$|z_2^2-z_1^2|+|z_3^2-z_2^2|+…|z_1^2-z_{10}^2|\leq 4\pi.$$

What I understand: $$z$$ is a unimodular complex number with argument $$\theta_1$$. The expression $$z_k=z_{k-1}e^{i\theta_k}$$ just indicates that rotating $$z_{k-1}$$ anticlockwise by $$\theta_{k}$$ gives $$z_k$$. Also, $$\theta_1+\theta_2+ \theta_3+… + \theta_{10}=2\pi$$ tells us that $$z_{10}$$ is the complex number $$1+0\cdot i$$. (X axis rotated on to itself). So I get a figure: Now, statement P refers to just the perimeter of the polygon $$z_1z_2…z_{10}z_1$$, which is less than the circumference of the circle which is $$2\pi$$.

I have a problem with Q. My solution is to think that the $$z_i^2$$‘s are just all unimodular complex numbers with all the angles $$\theta_i$$ doubled. Thus the as we go from $$z_1^2$$ till $$z_{10}^2$$, we span the whole unit circle twice, ending at $$z_{10}^2=1.$$ So the total length of the sides of this (?? Double polygon?) is less than twice the circumference of the circle which is $$4\pi$$.

I’m a bit uncomfortable with the justification of Q. I think I can prove that the perimeter of a polygon is less than the circumference of its circumcircle (I haven’t attempted it yet)(#). This is a multiple choice question, so no extremely rigorous proofs are required.

Is it okay? How would you make the justification of Q rigorous? I’d also appreciate it if anyone could post an algebraic proof here.

EDIT: I managed to prove (#).
I wrote the perimeter as $$\displaystyle P=2R\sum_{k=1}^{10} \sin\frac{\theta_k}{2}.$$

Now, by Jensen’s inequality, since $$\sin x$$ is a concave function in $$[0,\frac12\pi]$$, we have

$$\displaystyle \frac{1}{10}\sum_{k=1}^{10} \sin\frac{\theta_k}{2}\leq \sin\left(\frac{1}{10} \sum_{k=1}^{10} \frac{\theta_k}{2}\right)=\sin \frac{\pi}{10}=\frac{\sqrt 5-1}{4}$$ so that $$\displaystyle P=2R\sum_{k=1}^{10} \sin\frac{\theta_k}{2}\leq 5(\sqrt 5-1)R\lt 2\pi R= Circumference.$$

• This is more precise that your previous question yesterday... Jun 28, 2022 at 9:34
• I know the term "unimodular" seems as a good name for complex numbers of modulus $1$, but that's not the meaning of "unimodular" in Mathematics. The usual name is "unit complex numbers". Jun 28, 2022 at 9:39
• Here is a related question you may find interesting Jun 28, 2022 at 17:54
• YESSS!!!!! Thank you @EthakkaappamwithChai !!!!!! OwO Jun 28, 2022 at 18:26
• @EthakkaappamwithChai since you post these kinds of answers quite regularly, would you mind taking a look at math.stackexchange.com/questions/4481438/…? Your kind of answers were the exact thing I was referring to. Jun 28, 2022 at 18:49

I’m a bit uncomfortable with the justification of Q

For a different hint: $$\;|z_2^2-z_1^2| = |z_2+z_1| \cdot |z_2 - z_1| \leq \left(|z_2| + |z_1|\right) \cdot |z_2-z_1| = 2\, |z_2-z_1|\,$$.

the perimeter of a polygon is less than the circumference of its circumcircle

Each side is smaller than the arc subtended, because the shortest distance between two points is the line segment between them.

• Nice! Thanks and +1 Jun 28, 2022 at 17:16
• Hello sir/ma’am @dxiv could you please see my edit and help me to prove $5(\sqrt 5-1)\lt 2\pi$? Like, from the approximate values it’s obvious, but still… Jun 28, 2022 at 17:28
• @insipidintegrator More generally, you get $\,P \le 2R \,n \sin \frac{\pi}{n}\,$ from Jensen's inequality, then see Proving $n\sin(\frac{\pi}{n})<\pi<n\tan(\frac{\pi}{n})$ ; obtaining results from it..
– dxiv
Jun 28, 2022 at 17:34
• Thanks. Problem solved. Jun 28, 2022 at 17:39