Let theta_1, theta_2, ldots, theta_{10} be po | vedclass

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(C) Given $|z_1| = |z_2| = \ldots = |z_{10}| = 1$. The distance between two points $z_a$ and $z_b$ on the unit circle is $|z_a - z_b| = 2 \sin(\frac{\

Vedclass · Accepted Answer

both $P$ and $Q$ are $TRUE$

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(C) Given $|z_1| = |z_2| = \ldots = |z_{10}| = 1$. The distance between two points $z_a$ and $z_b$ on the unit circle is $|z_a - z_b| = 2 \sin(\frac{\Delta 	heta}{2})$,where $\Delta 	heta$ is the angle between them. Since $\sin(x) \leq x$ for $x \geq 0$,we have $|z_k - z_{k-1}| = 2 \sin(\frac{	heta_k}{2}) \leq 2(\frac{	heta_k}{2}) = 	heta_k$. Summing these,$\sum_{k=1}^{10} |z_{k+1} - z_k| \leq \sum_{k=1}^{10} 	heta_k = 2 \pi$ (where $z_{11} = z_1$). Thus,$P$ is $TRUE$. For $Q$,let $w_k = z_k^2 = e^{i 2 \phi_k}$,where $\phi_k = \sum_{j=1}^k 	heta_j$. The angle between $w_k$ and $w_{k-1}$ is $2 	heta_k$. Similarly,$|w_k - w_{k-1}| = |z_k^2 - z_{k-1}^2| = 2 \sin(\frac{2 	heta_k}{2}) = 2 \sin(	heta_k) \leq 2 	heta_k$. Summing these,$\sum |z_k^2 - z_{k-1}^2| \leq \sum 2 	heta_k = 2(2 \pi) = 4 \pi$. Thus,$Q$ is $TRUE$.

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