Problem 14 - Advanced Engineering Mathematics

Question

Verify the given result. \(\cos \left(\frac{\pi}{2}+i \ln 2\right)=-\frac{3}{4} i\)

Step-by-Step Solution

Answer

The given result \(-\frac{3}{4} i\) is verified to be correct.

Step 1: Recall the Formula for Complex Cosine

Recall the expression for cosine of a complex number \(z = x + yi\). For \(z = \frac{\pi}{2} + i\ln 2\), use the formula: \[\cos(z) = \cos(x + yi) = \cos(x)\cosh(y) - i\sin(x)\sinh(y)\] where \(\cosh(y)\) and \(\sinh(y)\) are the hyperbolic cosine and sine respectively.

Step 2: Substitute the Values of x and y

Given \(z = \frac{\pi}{2} + i\ln 2\), so \(x = \frac{\pi}{2}\) and \(y = \ln 2\). Using the formula from Step 1, substitute \(x\) and \(y\) into: \[\cos\left(\frac{\pi}{2} + i\ln 2\right) = \cos\left(\frac{\pi}{2}\right)\cosh(\ln 2) - i\sin\left(\frac{\pi}{2}\right)\sinh(\ln 2)\]

Step 3: Calculate \(\cos\left(\frac{\pi}{2}\right)\) and \(\sin\left(\frac{\pi}{2}\right)\)

We have \(\cos\left(\frac{\pi}{2}\right) = 0\) and \(\sin\left(\frac{\pi}{2}\right) = 1\). Substitute these values into the previous expression: \[\cos\left(\frac{\pi}{2} + i\ln 2\right) = 0 \cdot \cosh(\ln 2) - i \cdot 1 \cdot \sinh(\ln 2)\] Simplifying, we get: \[\cos\left(\frac{\pi}{2} + i\ln 2\right) = -i\sinh(\ln 2)\]

Step 4: Evaluate \(\sinh(\ln 2)\)

Recall that \(\sinh(y) = \frac{e^y - e^{-y}}{2}\). For \(y = \ln 2\), we have \[\sinh(\ln 2) = \frac{e^{\ln 2} - e^{-\ln 2}}{2} = \frac{2 - \frac{1}{2}}{2}\] Simplify further: \[= \frac{\frac{4}{2} - \frac{1}{2}}{2} = \frac{3}{4}\] Therefore, \(\sinh(\ln 2) = \frac{3}{4}\).

Step 5: Substitute \(\sinh(\ln 2)\) into the Expression

Substitute the calculated \(\sinh(\ln 2)\) into the expression from Step 3: \[\cos\left(\frac{\pi}{2} + i\ln 2\right) = -i \cdot \frac{3}{4}\] Hence, \(\cos\left(\frac{\pi}{2} + i\ln 2\right) = -\frac{3}{4}i\), which matches the given result.

Key Concepts

Complex Cosine

The complex cosine function involves calculating the cosine of a complex number. A complex number is generally expressed as \(z = x + yi\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit. To compute the cosine of a complex number, we use the formula:

• \( \cos(z) = \cos(x + yi) = \cos(x)\cosh(y) - i\sin(x)\sinh(y) \)

Here, \( \cosh(y) \) and \( \sinh(y) \) represent the hyperbolic cosine and sine functions, respectively. This formula explains how complex trigonometric functions can be broken down into their real and imaginary components using hyperbolic functions. When you substitute specific values for \(x\) and \(y\), you can solve for the complex cosine, as seen in the given exercise.

Hyperbolic Functions

Hyperbolic functions like \( \cosh(y) \) and \( \sinh(y) \) are analogs to trigonometric functions but for the hyperbola instead of a circle. They are essential in dealing with complex numbers and have several applications in physics and mathematics.

• \( \cosh(y) = \frac{e^y + e^{-y}}{2} \)
• \( \sinh(y) = \frac{e^y - e^{-y}}{2} \)

In the context of this exercise, the hyperbolic sine \( \sinh(y) \) is used to compute the imaginary part of the complex cosine function. By evaluating \( \sinh(\ln 2) \), we determine its value and thereby find part of the solution to the complex cosine expression. Understanding these functions is key to working with complex numbers.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental tools in mathematics. They are defined on the unit circle and extend to complex numbers through Euler's formula and other trigonometric identities.

• \( \cos(\frac{\pi}{2}) = 0 \)
• \( \sin(\frac{\pi}{2}) = 1 \)

The above values are specifically used to simplify the complex cosine calculation in our exercise. These functions relate angles to ratios of sides in right triangles and are extended for complex numbers using hyperbolic functions. In the exercise, the real part \( \cos(\frac{\pi}{2}) \) becomes zero, simplifying the expression. The imaginary part involving \( \sin(\frac{\pi}{2}) \) captures the hyperbolic component, leading to the verification of the provided result.

Exponential Function

The exponential function \( e^x \) is a crucial mathematical function that remains unique in how it behaves under differentiation and integration. It is frequently used in various mathematical fields and plays a key role in complex number calculations.When dealing with hyperbolic functions:

• \( e^{\ln a} = a \)
• \( e^{-\ln a} = \frac{1}{a} \)

In the problem, when calculating \( \sinh(\ln 2) \), the exponential function simplifies to provide the fractional components of \( \frac{3}{4} \), ensuring accuracy in computing hyperbolic sine. The exponential function bridges the logarithm, complex arithmetic, and trigonometric identities, providing a powerful tool in solving and verifying expressions involving complex numbers.