Problem 46 - Advanced Engineering Mathematics

Question

If \(z_{1}=-i\) and \(z_{2}=i\) then \\[ \operatorname{Ln}\left(z_{1} / z_{2}\right)=\operatorname{Ln}(-1)=\pi i, \quad \text { whereas } \quad \operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}=-\frac{\pi}{2} i-\frac{\pi}{2} i=-\pi i \\]

Step-by-Step Solution

Answer

\( \operatorname{Ln}(z_1/z_2) = \pi i \) and \( \operatorname{Ln} z_1 - \operatorname{Ln} z_2 = -\pi i \).

Step 1: Define the Complex Logarithm

The complex logarithm function is denoted by \( \operatorname{Ln} z \) and is given by \( \operatorname{Ln} z = \ln |z| + i \arg(z) \), where \( |z| \) is the modulus of \( z \) and \( \arg(z) \) is the argument of \( z \).

Step 2: Find Moduli and Arguments of \( z_1 \) and \( z_2 \)

For \( z_1 = -i \), \( |z_1| = 1 \) and \( \arg(z_1) = -\frac{\pi}{2} \). For \( z_2 = i \), \( |z_2| = 1 \) and \( \arg(z_2) = \frac{\pi}{2} \).

Step 3: Calculate \( \operatorname{Ln} z_1 \) and \( \operatorname{Ln} z_2 \)

Using the formula, \( \operatorname{Ln}(-i) = \ln 1 + i(-\frac{\pi}{2}) = -\frac{\pi}{2} i \). Similarly, \( \operatorname{Ln}(i) = \ln 1 + i(\frac{\pi}{2}) = \frac{\pi}{2} i \).

Step 4: Compute \( \operatorname{Ln} z_1 - \operatorname{Ln} z_2 \)

Subtract the results: \( -\frac{\pi}{2} i - \frac{\pi}{2} i = -\pi i \).

Step 5: Calculate \( z_1 / z_2 \)

\( \frac{z_1}{z_2} = \frac{-i}{i} = -1 \).

Step 6: Find \( \operatorname{Ln}(z_1/z_2) \)

Since \( z_1 / z_2 = -1 \), we have \( \operatorname{Ln}(-1) = \ln 1 + i\pi = \pi i \).

Step 7: Conclusion and Verification

You have shown through these calculations that indeed, \( \operatorname{Ln}(z_1 / z_2) = \pi i \) and \( \operatorname{Ln} z_1 - \operatorname{Ln} z_2 = -\pi i \), which matches the given information.

Key Concepts

Modulus

In the world of complex numbers, the modulus of a complex number can be thought of as its "distance" from the origin on the complex plane. For any complex number, given in the form of \(a + bi\), the modulus can be found using the formula:

• \(|z| = \sqrt{a^2 + b^2}\)

This gives a non-negative real number that represents the size or length of the complex number vector. For example, in the exercise, for both \( z_1 = -i \) and \( z_2 = i \), their moduli are calculated as follows:

• \(|z_1| = 1\)
• \(|z_2| = 1\)

Both have a modulus of 1 because they lie one unit away from the origin on the imaginary axis. Modulus plays a crucial role in the complex logarithm as it helps in transforming the polar representation of complex numbers into the logarithmic form.

Argument of Complex Number

A complex number's argument is the angle formed with the positive real axis on the complex plane. It provides direction to the vector represented by the complex number. To find the argument, we usually use the arctangent function. Given a complex number \( z = a + bi \), the argument is calculated as:

• \( \arg(z) = \tan^{-1}\left(\frac{b}{a}\right)\)

In practice, the arguments of complex numbers are often measured in radians.For instance, in the exercise:

• \( \arg(z_1) = -\frac{\pi}{2} \)
• \( \arg(z_2) = \frac{\pi}{2} \)

These reflect that \( z_1 \) is rotated 90 degrees clockwise from the positive imaginary axis, while \( z_2 \) is rotated 90 degrees counterclockwise. The understanding of arguments is fundamental when dealing with complex logarithms, as they directly influence the imaginary part of the logarithm.

Complex Division

Complex division involves dividing one complex number by another and is analogous to multiplying by the reciprocal of the divisor. To divide two complex numbers \( z_1 = a + bi \) and \( z_2 = c + di \), we employ the formula:

• \( \frac{z_1}{z_2} = \frac{(a + bi)(c - di)}{c^2 + d^2} \)

This results in a new complex number where the denominator is guaranteed to be real and positive, making further calculations straightforward.In the provided example:

• \( \frac{z_1}{z_2} = \frac{-i}{i} = -1 \)

This simplification shows the beauty of complex division. The division sub-problems are resolved into simpler real-number operations, indicating clear solutions. In this context, complex division helps verify the log properties by ensuring the division outcome aligns with the solved arguments. It ensures consistency when applying complex logarithmic functions, which is crucial for multiplying and dividing complex exponentials.