Let θ1, θ2, ..., θ10 be positive valued angles (in radian) such that θ1 + θ2 + ... + θ10 = 2π. Defin

Q: Wait, why can we just say $|z_k - z_{k-1}| \le \theta_k$? Is this always true?

That is a brilliant question! In the unit circle, the chord length between two points with an angular separation of $\theta$ is $2\sin(\theta/2)$. We know from trigonometry that for any non-negative angle $\theta$, $\sin(\theta/2) \le \theta/2$. Multiplying both sides by $2$, we get $2\sin(\theta/2) \le \theta$. Thus, the chord length is indeed always less than or equal to the arc length. It is a beautiful, fundamental geometric truth!

Q: Why does squaring the complex numbers change the arc length to $2\theta_k$?

Think about the polar form: $z_k = e^{i\phi_k}$, where $\phi_k = \sum_{j=1}^k \theta_j$. When we square $z_k$, we get $z_k^2 = e^{i(2\phi_k)}$. The new angle is $2\phi_k$. The difference in angle between $z_k^2$ and $z_{k-1}^2$ is $2\phi_k - 2\phi_{k-1} = 2(\phi_k - \phi_{k-1}) = 2\theta_k$. So, the arc length between the new points is exactly double the original arc length!

Q: What if the sum of angles was greater than $2\pi$? Would the inequality still hold?

If the sum of angles exceeded $2\pi$, the points would wrap around the circle more than once. The inequality $|z_k - z_{k-1}| \le \theta_k$ would still hold for each individual segment, but the sum of the chord lengths would no longer be bounded by $2\pi$. The constraint $\sum \theta_k = 2\pi$ is crucial because it defines the total 'budget' of arc length we have to work with!

Question

Let $	heta_1, 	heta_2, \dots, 	heta_{10}$ be positive valued angles (in radian) such that $	heta_1 + 	heta_2 + \dots + 	heta_{10} = 2\pi$. Define the complex numbers $z_1 = e^{i	heta_1}$, $z_k = z_{k-1}e^{i	heta_k}$ for $k = 2, 3, \dots, 10$, where $i = \sqrt{-1}$. Consider the statements P and Q given below : 
P : $|z_2 - z_1| + |z_3 - z_2| + \dots + |z_{10} - z_9| + |z_1 - z_{10}| \le 2\pi$ 
Q : $|z_2^2 - z_1^2| + |z_3^2 - z_2^2| + \dots + |z_{10}^2 - z_9^2| + |z_1^2 - z_{10}^2| \le 4\pi$ 
Then,

Options:

(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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler, to the beautiful world of complex numbers. Often, when we see expressions like $|z_k - z_{k-1}|$, our instinct is to reach for the algebraic toolkit: expanding $z_k$ into its real and imaginary parts, using the distance formula, and getting lost in a forest of trigonometric identities.

But today, we are going to do something different. We are going to look at the geometry. Imagine you are standing at the center of a unit circle in the complex plane.

You have ten points, $z_1, z_2, \dots, z_{10}$, all dancing on the circumference. Each point $z_k$ is defined by the sum of angles $	heta_1, 	heta_2, \dots, 	heta_k$. This means:

$$z_k = e^{i(	heta_1 + \dots + 	heta_k)}$$

Since the magnitude of $e^{i\phi}$ is always $1$, every single one of these points lies exactly on the unit circle.

### The Chord and the Arc

Now, look at the expression in Statement P: $|z_2 - z_1| + |z_3 - z_2| + \dots + |z_1 - z_{10}|$. In the complex plane, the modulus of the difference between two complex numbers is the distance between them.

If you connect $z_{k-1}$ and $z_k$ with a straight line, you have drawn a chord of the circle. The length of this chord is $|z_k - z_{k-1}|$.

But there is another way to get from $z_{k-1}$ to $z_k$: you could walk along the curve of the circle. This is the arc length. For a unit circle, the arc length between two points is simply the angle between them, which is $	heta_k$.

Here is the profound insight: the shortest distance between two points is a straight line. Therefore, the chord length must always be less than or equal to the arc length. Mathematically:

$$|z_k - z_{k-1}| \le 	heta_k$$

This is the key that unlocks the entire problem.

### Proving Statement P

If we sum this inequality for all $k$ from $1$ to $10$, we get:

$$\sum_{k=1}^{10} |z_k - z_{k-1}| \le \sum_{k=1}^{10} 	heta_k$$

We are given that the sum of all angles is $2\pi$. Therefore, the sum of the chord lengths is less than or equal to $2\pi$. **Statement P is true!** It is not just true; it is elegantly, geometrically inevitable.

### The Transformation: Statement Q

Now, let us tackle Statement Q. We are looking at the squares of these complex numbers: $w_k = z_k^2$. When you square a complex number on the unit circle, you are doubling its angle.

If $z_k$ had an angle of $\phi_k$, then $w_k$ has an angle of $2\phi_k$. The points $w_k$ are still on the unit circle, but they are moving twice as fast.

The distance between $w_k$ and $w_{k-1}$ is the chord length of the new points. The angle between them is $2	heta_k$. Using the same geometric logic, the chord length $|w_k - w_{k-1}|$ must be less than or equal to the new arc length, which is $2	heta_k$:

$$|z_k^2 - z_{k-1}^2| \le 2	heta_k$$

Summing this over all $k$, we get:

$$\sum |z_k^2 - z_{k-1}^2| \le \sum 2	heta_k = 2 \sum 	heta_k = 2(2\pi) = 4\pi$$

**Statement Q is also true!**

### The Takeaway

We have navigated through the problem without a single messy algebraic expansion. We used the power of geometric intuition to see the truth hidden behind the symbols.

Remember this: whenever you see complex numbers on a circle, stop and visualize the geometry. The chords, the arcs, and the angles are your best friends. You have mastered this problem, not by brute force, but by understanding the soul of the mathematics.

Select Answer:

Visualized Solution

$z_{k} = e^{i (θ_{1} + \dots + θ_{k})}$

$z_{10} = e^{i (2 π)}$

Visualizing $z_{k}$ and $θ_{k}$

Chord Length $= ∣ z_{k} - z_{k - 1} ∣$

Arc Length $= θ_{k}$

Chord $<$ Arc

$\sum ∣ z_{k} - z_{k - 1} ∣ \leq 2 π$

$w_{k} = z_{k}^{2}$

$∣ z_{k}^{2} - z_{k - 1}^{2} ∣ \leq 2 θ_{k}$

$\sum ∣ z_{k}^{2} - z_{k - 1}^{2} ∣ \leq 4 π$

Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Chord and the Arc

Proving Statement P

The Transformation

The Takeaway

Similar Questions

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

Comprehension Passage

Area of $S =$

$min_{z \in S} ∣1 - 3 i - z ∣ =$

Let $C$ be the set of all complex numbers. Let $S_{1} = {z \in C : ∣ z - 2∣ \leq 1}$ and $S_{2} = {z \in C : z (1 + i) + \overset{z}{ˉ} (1 - i) \geq 4}$ . Then, the maximum value of $∣ z - \frac{5}{2} ∣^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to :

Match the statements in Column I with those in Column II. [Note : Here $z$ takes values in the complex plane and $I m z$ and $R ez$ denote , respectively, the imaginary part and the real part of $z$ . ]

List-I

List-II

For a non-zero complex number $z$ , let $ar g (z)$ denote the principal argument with $- π < ar g (z) \leq π$ . Then, which of the following statement(s) is (are) FALSE ?

Let $O$ be the origin and $A$ be the point $z_{1} = 1 + 2 i$ . If $B$ is the point $z_{2}$ , $Re (z_{2}) < 0$ , such that $O A B$ is a right angled isosceles triangle with $O B$ as hypotenuse, then which of the following is NOT true ?

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } zk​=ei(θ1​+⋯+θk​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z10​=ei(2π)

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } zk​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θk​

Chord Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } =∣zk​−zk−1​∣

Arc Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } =θk​

Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } < Arc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑∣zk​−zk−1​∣≤2π

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } wk​=zk2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣zk2​−zk−12​∣≤2θk​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑∣zk2​−zk−12​∣≤4π