Problem 132 - A Complete Resource Book in Mathematics for JEE Main

Question

\(e^{2 m i \cot ^{-1} p} \cdot\left(\frac{p i+1}{p i-1}\right)^{m}=\) (A) 0 (B) 1 (C) \(-1\) (D) None of these

Step-by-Step Solution

Answer

The expression simplifies to 1.

Step 1: Understanding the Problem

We are given the expression \(e^{2 m i \cot^{-1} p} \cdot\left(\frac{p i+1}{p i-1}\right)^{m}\). We need to find its value or simplify it to match one of the provided options.

Step 2: Simplifying the cotangent inverse part

Recall that \( \cot^{-1} p \) is the inverse cotangent function. The expression \(e^{2 m i \cot^{-1} p}\) involves the exponential function, which can also be written in terms of complex trigonometric functions.

Step 3: Expressing Exponents using Cotangent Identity

We know that \( e^{i \theta} = \cos \theta + i \sin \theta \). If \( \theta = \cot^{-1} p \), then \( \cos(\theta) = \frac{1}{\sqrt{1+p^2}} \) and \( \sin(\theta) = \frac{p}{\sqrt{1+p^2}} \). Hence, \( e^{i 2m \cot^{-1} p} = \left(\frac{1}{\sqrt{1+p^2}} + i \frac{p}{\sqrt{1+p^2}}\right)^{2m}. \)

Step 4: Calculate \(\left(\frac{p i+1}{p i-1}\right)^{m}\)

The expression \( \frac{p i+1}{p i-1} \) inside the power can be simplified further. Multiply the numerator and the denominator by the conjugate of the denominator: \( \frac{(p i + 1)(-p i - 1)}{(p i - 1)(-p i - 1)} \). This simplifies to \( e^{i \cot^{-1} p} \) as well.

Step 5: Combine and Simplify the Expressions

Now we have \( \left(e^{i \cdot 2m \cdot \cot^{-1} p} \right) \cdot \left(e^{i \cdot \cot^{-1} p} \right)^{m} = e^{i \cdot (2m \cot^{-1} p + m \cot^{-1} p)} \).This further simplifies to \(e^{i \cdot 3m \cot^{-1} p}\).

Step 6: Simplify Further Using Periodicity

The simplified expression \(e^{i \cdot 3m \cot^{-1} p }\) will evaluate to 1 because the exponent is essentially just adding multiple of the period length of complex exponentials cycle.

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics, particularly important when dealing with complex numbers and their applications. A common form of an exponential function with a complex number is represented as \( e^{ix} \), where \( x \) is a real number and \( i \) is the imaginary unit. This can be expressed using Euler's formula:

• \( e^{ix} = \cos(x) + i\sin(x) \)

This formula allows us to link complex exponentials with trigonometric functions, making it easier to simplify and manipulate such expressions.

In the context of manipulating complex expressions, understanding exponential functions helps in converting trigonometric identities, simplifying complex powers, and integrating inverse trigonometric functions. When working with exponential functions, especially in complex numbers, recognizing cyclical behavior due to the periodicity of trigonometric functions is crucial, as this can lead to simplifications, such as reducing the powers to smaller, equivalent values to evaluate expressions effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions like \( \cot^{-1}(x) \) allow us to determine angles based on their trigonometric values. These functions are essential in calculus and complex analysis because they provide solutions to equations involving trigonometric identities. In the original exercise, \( \cot^{-1}(p) \) plays a crucial role, helping us link trigonometric expressions with the given exponential form.

Inverse trigonometric functions are often expressed in complex exponential forms for simplification. For example, with \( \theta = \cot^{-1}(p) \), this expression translates well into an exponential form such as \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \). Knowledge of how \( \cos(\theta) \) and \( \sin(\theta) \) relate to properties of \( \tan^{-1} \), \( \cot^{-1} \), and their respective triangles is fundamental for such transformations. Using these transformations is crucial when combining trigonometric equations and manipulating them within complex expressions.

Algebraic Manipulations

Algebraic manipulations involve techniques used to simplify expressions, solve equations, and transform functions. In the context of complex numbers and exponential functions, these manipulations often include multiplying by conjugates, distributing exponents, and using trigonometric identities.

In the example we reviewed, simplifying \( \left( \frac{p i + 1}{p i - 1} \right)^m \) involved multiplying the fraction by the conjugate of the denominator. This step simplifies the fraction into a familiar trigonometric form, proving extremely useful in conjunction with exponential transformations.

• Make equations easier to handle.
• Reveal periodic behaviors in complex numbers.

These steps are well-valued in algebra because they provide a pathway from a complicated expression to a simplified or standard form that is easier to evaluate and understand. Through these manipulations, we can perform operations such as finding equivalent expressions and pairing them with trigonometric identities to simplify higher powers or expressions involving complex numbers.