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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:

enter image description here

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

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1 Answer 1

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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.

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