Question

Let ${\theta }_1,{\theta }_2,\dots ..,{\theta }_{10}$ be positive valued angles (in radian) such that ${\theta }_1+{\theta }_2+\dots ..+{\theta }_{10}=2\pi$. Define the complex numbers $z_1=e^{i{\theta }_1},z_k=z_{k-1}{e}^{i{\theta }_k}$ for $k=2,3,\dots \dots ,10$, where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P:\left|z_2-z_1\right|+\left|z_3-z_2\right|+\dots .+\left|z_{10}-z_9\right|+\left|z_1-z_{10}\right|\le 2\pi$
$Q:\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\dots ..+\left|z_{10}^2-z_9^2\right|+\left|z_1^2-z_{10}^2\right|\le 4\pi$

• A. P is TRUE and Q is FALSE
• B. Q is TRUE and P is FALSE
• C. both P and Q are TRUE
• D. both P and Q are FALSE

Answer 1

Answer: (C)

$\because z_1=e^{i{\theta }_1}$
So, $z_2=e^{i\left({\theta }_1+{\theta }_2\right)}$
$z_3=e^{i\left({\theta }_1+{\theta }_2+{\theta }_3\right)}$
:
$z_{10}=e^{i\left({\theta }_1+{\theta }_2+\dots \dots +{\theta }_{10}\right)}=e^{i(2\pi )}$
Sum of all the chord length $<$ Circumference
So, $\sum \left|z_2-z_1\right|\le 2\pi$
Also, $2\left|z_2-z_1\right|\ge \left|z_2^2-z_1^2\right|$
Hence, $2\left(\left|z_2-z_1\right|+\dots ..+\left|z_{10}-z_1\right|\right)\le 2(2\pi )=4\pi$
So for, we have $P\le 2\pi$ and $Q\le 4\pi$