Let θ 1 , θ 2 , ldots . ., θ 10 be positive valued angles (in radian) such that θ 1 +θ 2 +ldots . .+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, ldots . ., 10 , where i=√(-1) . Consider the statements P and Q given below:

Mathematics · Complex Numbers

JEE Advanced 2021 — Paper 1 — Question 22

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:

Option A:
P is TRUE and Q is FALSE
Option B:
Q is TRUE and P is FALSE
Option C:
both P and Q are TRUE
Correct
Option D:
both $P$ and $Q$ are FALSE

Answer: C

Step-by-step solution

$\because \mathrm{z}_{1}=\mathrm{e}^{\mathrm{i} \theta_{1}}$ So, $z_{2}=e^{i\left(\theta_{1}+\theta_{2}\right)}$ $z_{3}=e^{i\left(\theta_{1}+\theta_{2}+\theta_{3}\right)}$ $\vdots$ $z_{10}=e^{i\left(\theta_{1}+\theta_{2}+\ldots . .+\theta_{10}\right)}=e^{i(2 \pi)}$

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 $\mathrm{P} \leq 2 \pi$ and $\mathrm{Q} \leq 4 \pi$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Geometry of Complex Numbers

For any complex number w= c+ id, let arg ( w) in(-pi, pi] , where i=√(-1) . Let alpha and beta be real numbers such that for all complex…

Answer: C

Step-by-step solution

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