JEE Advanced 2021 Paper 1 Q04 Mathematics Algebra Complex Numbers Hard

JEE Advanced 2021 Paper 1 · Q04 · Complex Numbers

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1 + \theta_2 + \cdots + \theta_{10} = 2\pi$. Define the complex numbers $z_1 = e^{i\theta_1}$, $z_k = z_{k-1} e^{i\theta_k}$ for $k = 2, 3, \ldots, 10$, where $i = \sqrt{-1}$. Consider the statements $P$ and $Q$ given below: $P: |z_2 - z_1| + |z_3 - z_2| + \cdots + |z_{10} - z_9| + |z_1 - z_{10}| \leq 2\pi$ $Q: |z_2^2 - z_1^2| + |z_3^2 - z_2^2| + \cdots + |z_{10}^2 - z_9^2| + |z_1^2 - z_{10}^2| \leq 4\pi$ Then,

  1. A. P is TRUE and Q is FALSE
  2. B. Q is TRUE and P is FALSE
  3. C. both P and Q are TRUE
  4. D. both P and Q are FALSE
Reveal answer + step-by-step solution

Correct answer:C

Solution

Each $z_k = e^{i(\theta_1+\theta_2+\cdots+\theta_k)}$ lies on the unit circle. The points $z_1,z_2,\ldots,z_{10}$ visit the unit circle and return to $z_1$ (since the total angle is $2\pi$). The sum $\sum|z_{k+1}-z_k|$ is the perimeter of an inscribed polygon, which is at most the circumference $2\pi$. Hence $P$ is TRUE. For $Q$, $z_k^2=e^{2i(\theta_1+\cdots+\theta_k)}$ traces the unit circle while the total angle doubles to $4\pi$ (the points wind around twice). Thus $\sum|z_{k+1}^2-z_k^2|\le 2(2\pi)=4\pi$. So $Q$ is also TRUE. Answer: (C).

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