Question

Let $\theta_{1}, \theta_{2}, \ldots . ., \theta_{10}$ be positive valued angles (in radian) such that $\theta_{1}+\theta_{2}+\ldots . .+\theta_{10}=2 \pi$. Define the complex numbers $z_{1}=e^{i \theta_{1}}, z_{k}=z_{k-1} e^{i \theta_{k}}$ for $k=2,3, \ldots \ldots, 10$, 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|+\ldots .+\left|z_{10}-z_{9}\right|+\left|z_{1}-z_{10}\right| \leq 2 \pi $
$Q:\left|z_{2}^{2}-z_{1}^{2}\right|+\left|z_{3}^{2}-z_{2}^{2}\right|+\ldots . .+\left|z_{10}^{2}-z_{9}^{2}\right|+\left|z_{1}^{2}-z_{10}^{2}\right| \leq 4 \pi$

• 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, $z_{2}=e^{i\left(\theta_{1}+\theta_{2}\right)}$
$z_{3}=e^{i\left(\theta_{1}+\theta_{2}+\theta_{3}\right)}$
:
$z _{10}= e ^{ i \left(\theta_{1}+\theta_{2}+\ldots \ldots+\theta_{10}\right)}= e ^{ i (2 \pi)}$
Sum of all the chord length $ < $ Circumference
So, $\sum\left|z_{2}-z_{1}\right| \leq 2 \pi$
Also, $2\left|z_{2}-z_{1}\right| \geq\left|z_{2}^{2}-z_{1}^{2}\right|$
Hence, $2\left(\left|z_{2}-z_{1}\right|+\ldots . .+\left|z_{10}-z_{1}\right|\right) \leq 2(2 \pi)=4 \pi$
So for, we have $P \leq 2 \pi$ and $Q \leq 4 \pi$