JEE Advanced · Mathematics · 3. 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 $i = \sqrt{- 1}$ . Consider the statements $P$ and $Q$ given below:
$P:\left|z_2-z_1\right|+\left|z_3-z_2\right|+\cdots+\left|z_{10}-z_9\right|$ $+\left|z_1-z_{10}\right| \leq 2 \pi $
$ Q:\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\cdots+\left|z_{10}^2-z_9^2\right|$ $+\left|z_1^2-z_{10}^2\right| \leq 4 \pi$

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

Verified Solution

Answer & Solution

Correct Answer

(C) both $P$ and $Q$ are TRUE

Step-by-step Solution

Detailed explanation

Given,

θ_{1} + θ_{2} + . . . . . + θ_{10} = 2 π

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

Now,

z_{k} = z_{k - 1} e^{i θ_{k}}

\Rightarrow z_{2} = z_{1} e^{i θ_{1}}, z_{3} = z_{2} e^{i θ_{2}}, \dots \dots

As,

z_{1} = e^{i θ_{1}}

\Rightarrow |z_{1}| = 1

z_{2} = z_{1} e^{_{i θ}}

\Rightarrow |z_{2}| = |z_{1}| = 1

Similarly

|z_{1}| = |z_{2}| \dots = |z_{k}| = 1

\Rightarrow z_{1}, z_{2}, \dots, z_{k}

lies on a circle of unit radius.
We know,

|z_{k} - z_{k - 1}|

represents a line segment joining

z_{k} & z_{k - 1}

.
Both

z_{k} & z_{k - 1}

lies on a unit circle

Since

θ_{1} + θ_{2} + . . . . + θ_{10} = 2 π

We get

So,

|z_{2} - z_{1}|, |z_{3} - z_{2}|, \dots, |z_{1} - z_{10}|

are the sides of a decagon circumscribed by a circle of unit radius
We know, the sum of the length of sides of a decagon is less than the circumference of its circumcircle.
$\Rightarrow\left|z_2-z_1\right|+\left|z_3-z_2\right|+\cdots+\left|z_{10}-z_9\right|$ $+\left|z_1-z_{10}\right| \leq 2 \pi$
Hence,

P

is true.
Similarly,

{(z_{1})}^{2} = e^{i 2 θ_{1}}, {(z_{k})}^{2} = {(z_{k - 1})}^{2} e^{i 2 θ_{k}}, \dots \dots

|z_{2}^{2} - z_{1}^{2}| + |z_{3}^{2} - z_{2}^{2}| + \dots |z_{6}^{2} - z_{5}^{2}| \leq 2 π

Similarly,

|z_{7}^{2} - z_{8}^{2}| + \dots + |z_{1}^{2} - z_{10}^{2}| \leq 2 π

Adding these equations, we get
$\left|z_2^2-z_1^2\right|+\left|z_3^2-z_2^2\right|+\cdots+\left|z_{10}^2-z_9^2\right|$ $+~\left|z_1^2-z_{10}^2\right| \leq 4 \pi$
Hence,

Q

is true.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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