If |Z| ≤ 1, |W| ≤ 1, show that |Z - W|² ≤ (|Z| - |W|)² + (Arg Z - Arg W)².

Q: Why can't we just use the triangle inequality $|Z - W| \le |Z| + |W|$? It seems much simpler.

That is a fantastic question! The triangle inequality is indeed a powerful tool, but it provides a very loose upper bound. In this problem, we are looking for a specific, tighter bound that accounts for the angular separation between the two complex numbers. The triangle inequality ignores the angle entirely, whereas our target expression explicitly includes $(Arg Z - Arg W)^2$. We need a more surgical approach.

Q: Why do we use the identity $1 - \cos A = 2\sin^2(A/2)$? Is there another way?

You could try Taylor series expansions, but the half-angle identity is the most elegant path. It transforms the cosine term into a sine squared term. This is crucial because we have a standard inequality, $|\sin x| \le |x|$, which allows us to linearize the angular difference. Without this identity, we would be stuck with a trigonometric function that is difficult to bound linearly.

Q: Is the inequality $|\sin x| \le |x|$ always true? I feel like I might be assuming too much.

You are absolutely right to be cautious! This is a fundamental result from calculus. If you look at the graph of $y = \sin x$ and $y = x$ near the origin, the line $y = x$ is always above the sine curve for $x > 0$. It is a rigorous inequality that holds for all real $x$. Trust this tool; it is one of the most reliable bridges between trigonometry and algebra.

Question

If $|Z| \le 1, |W| \le 1$, show that $|Z - W|^2 \le (|Z| - |W|)^2 + (Arg Z - Arg W)^2$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of complex numbers is a journey into the Argand plane. When considering the expression $|Z - W|^2 \le (|Z| - |W|)^2 + (Arg Z - Arg W)^2$, we are essentially exploring the distance between two points, $Z$ and $W$, trapped within the unit circle.

The term $|Z - W|$ represents the straight-line distance between these points. We aim to understand how this distance is constrained by the differences in their magnitudes and their angular positions.

### The Law of Cosines: Our Geometric Compass

To solve this, we translate the complex numbers into polar coordinates. Let $Z = r_1 e^{i	heta_1}$ and $W = r_2 e^{i	heta_2}$, where $r_1, r_2$ are the magnitudes and $	heta_1, 	heta_2$ are the arguments.

These points form a triangle with the origin, where the sides are $r_1$, $r_2$, and the distance $|Z - W|$. The angle between the vectors is $	heta_1 - 	heta_2$. Applying the Law of Cosines, we obtain:

$$|Z - W|^2 = r_1^2 + r_2^2 - 2r_1r_2 \cos(	heta_1 - 	heta_2)$$

### The Art of Algebraic Manipulation

We want to relate this to the target expression $(|Z| - |W|)^2 + (Arg Z - Arg W)^2$, which expands to $(r_1 - r_2)^2 + (	heta_1 - 	heta_2)^2$. We manipulate our Law of Cosines equation by adding and subtracting $2r_1r_2$:

$$|Z - W|^2 = (r_1^2 + r_2^2 - 2r_1r_2) + 2r_1r_2 - 2r_1r_2 \cos(	heta_1 - 	heta_2)$$

Grouping the terms, we identify the square of the difference of magnitudes:

$$|Z - W|^2 = (r_1 - r_2)^2 + 2r_1r_2(1 - \cos(	heta_1 - 	heta_2))$$

### The Trigonometric Bridge

To simplify the remaining term, we utilize the half-angle identity $1 - \cos A = 2\sin^2(\frac{A}{2})$. Applying this to our expression yields:

$$|Z - W|^2 = (r_1 - r_2)^2 + 4r_1r_2 \sin^2\left(\frac{	heta_1 - 	heta_2}{2}ight)$$

This separation allows us to analyze the radial and angular components independently.

### The Calculus Leap

We transition to an inequality using the fundamental property $|\sin x| \le |x|$, which implies $\sin^2 x \le x^2$. Setting $x = \frac{	heta_1 - 	heta_2}{2}$, we find:

$$\sin^2\left(\frac{	heta_1 - 	heta_2}{2}ight) \le \frac{(	heta_1 - 	heta_2)^2}{4}$$

Substituting this into our equation, the factors of $4$ cancel out:

$$|Z - W|^2 \le (r_1 - r_2)^2 + 4r_1r_2 \cdot \frac{(	heta_1 - 	heta_2)^2}{4}$$

$$|Z - W|^2 \le (r_1 - r_2)^2 + r_1r_2(	heta_1 - 	heta_2)^2$$

### The Final Constraint

Given that $|Z| \le 1$ and $|W| \le 1$, we know $r_1 \le 1$ and $r_2 \le 1$, which implies $r_1r_2 \le 1$. Replacing $r_1r_2$ with $1$ maintains the inequality:

$$|Z - W|^2 \le (r_1 - r_2)^2 + 1 \cdot (	heta_1 - 	heta_2)^2$$

Substituting back the original notation, we arrive at the final result:

**$$|Z - W|^2 \le (|Z| - |W|)^2 + (Arg Z - Arg W)^2$$**

Animated Solution for Mathematics - Complex Numbers: If $∣ Z ∣ \leq 1, ∣ W ∣ \leq 1$ , show that $∣ Z - W ∣^{2} \leq (∣ Z ∣ - ∣ W ∣)^{2} + (A r g Z - A r g W)^{2}$ .

Visualized Solution

Visualizing the Complex Plane

Polar Representation

The Distance $∣ Z - W ∣$

Law of Cosines

Algebraic Manipulation

Forming the Square

Trigonometric Identity

The Sine Inequality

Bounding the Expression

Using Given Constraints

Final Substitution

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Law of Cosines

The Art of Algebraic Manipulation

The Trigonometric Bridge

The Calculus Leap

The Final Constraint

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