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 $k=2,3, \ldots, 10$, where $i=\sqrt{-1}$. Consider the statements P and Q given below: $P:|z_{2}-z_{1}|+|z_{3}-z_{2}|+\ldots+|z_{10}-z_{9}|+|z_{1}-z_{10}| \leq 2\pi$ $Q:|z_{2}^{2}-z_{1}^{2}|+|z_{3}^{2}-z_{2}^{2}|+\ldots+|z_{10}^{2}-z_{9}^{2}|+|z_{1}^{2}-z_{10}^{2}| \leq 4\pi$ Then,

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Complex Numbers

Let

θ_{1}, θ_{2}, \dots, θ_{10}

be positive valued angles (in radian) such that

θ_{1} + θ_{2} + \dots + θ_{10} = 2 π

. Define the complex numbers

z_{1} = e^{i θ_{1}}, z_{k} = z_{k - 1} e^{i θ_{k}}

for

k = 2, 3, \dots, 10

, where

i = - 1

. Consider the statements P and Q given below:

P : ∣ z_{2} - z_{1} ∣ + ∣ z_{3} - z_{2} ∣ + \dots + ∣ z_{10} - z_{9} ∣ + ∣ z_{1} - z_{10} ∣ \leq 2 π

Q : ∣ z_{2}^{2} - z_{1}^{2} ∣ + ∣ z_{3}^{2} - z_{2}^{2} ∣ + \dots + ∣ z_{10}^{2} - z_{9}^{2} ∣ + ∣ z_{1}^{2} - z_{10}^{2} ∣ \leq 4 π

Then,

Answer: (C)

All

z_{k}

lie on the unit circle:

∣ z_{1} ∣ = ∣ z_{2} ∣ = \dots = ∣ z_{10} ∣ = 1

. The angle between consecutive points is

θ_{k}

. For points on a circle, the chord length is less than the arc length:

∣ z_{k + 1} - z_{k} ∣ \leq θ_{k}

. Therefore,

∣ z_{2} - z_{1} ∣ + \dots + ∣ z_{1} - z_{10} ∣ \leq θ_{1} + θ_{2} + \dots + θ_{10} = 2 π

, so P is true. For Q, consider

z_{k}^{2} = e^{i 2 θ_{k}}

. Let

α_{k} = 2 θ_{k}

, so

z_{1}^{2} = e^{i α_{1}}

and

z_{k}^{2} = z_{k - 1}^{2} \cdot e^{i α_{k}}

. We have

α_{1} + α_{2} + \dots + α_{10} = 4 π

. By the same reasoning,

∣ z_{1}^{2} - z_{2}^{2} ∣ + \dots + ∣ z_{1}^{2} - z_{10}^{2} ∣ \leq 4 π

, so Q is also true.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option