Let $ heta_{1}, heta_{2}, \ldots . ., heta_{10}$ be positive valued angles (in radian) such tha

Question

Let $	heta_{1}, 	heta_{2}, \ldots . ., 	heta_{10}$ be positive valued angles (in radian) such that $	heta_{1}+	heta_{2}+\ldots . .+	heta_{10}=2 \pi$. Define the complex numbers $z_{1}=e^{i 	heta_{1}}, z_{k}=z_{k-1} e^{i 	heta_{k}}$ for $k=2,3, \ldots . ., 10$, where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below: P: $\left|z_{2}-z_{1}ight|+\left|z_{3}-z_{2}ight|+\ldots . .+\left|z_{10}-z_{9}ight|+\left|z_{1}-z_{10}ight| \leq 2 \pi$ $Q:\left|z_{2}{ }^{2}-z_{1}{ }^{2}ight|+\left|z_{3}{ }^{2}-z_{2}{ }^{2}ight|+\ldots . .+\left|z_{10}^{2}-z_{9}^{2}ight|+\left|z_{1}^{2}-z_{10}^{2}ight| \leq 4 \pi$ Then,\

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(c): We have, $\left|z_{2}-z_{1}ight| \leq 	heta_{1}$ $\left|z_{3}-z_{2}ight| \leq 	heta_{2}$\ $\left|z_{1}-z_{10}ight| \leq 	heta_{10}$\ Adding all these, we get\ $\left|z_{2}-z_{1}ight|+\left|z_{3}-z_{2}ight|+\ldots$ $+\left|z_{1}-z_{10}ight| \leq 2 \pi$\ Now, $\left|z_{2}{ }^{2}-z_{1}{ }^{2}ight|$\ $=\left|z_{2}-z_{1}ight|\left|z_{2}+z_{1}ight| \leq 2\left|z_{2}-z_{1}ight|$\ $\left[\because\left|z_{2}+z_{1}ight| \leq\left|z_{2}ight|+\left|z_{1}ight|=2ight]$\ We have,\ $\left|z_{2}^{2}-z_{1}^{2}ight|+\left|z_{3}^{2}-z_{2}^{2}ight|+\ldots .+\left|z_{1}^{2}-z_{10}^{2}ight| \leq 2 \cdot 2 \pi=4 \pi$\