Problem 24 - Advanced Engineering Mathematics

Question

Express \(\ln z\) in the form \(a+i b\). \(z=-e i\)

Step-by-Step Solution

Answer

\(\ln z = 1 - i\frac{\pi}{2}\)

Step 1: Express the complex number in polar form

The given complex number is \(z = -e i\). We can express any complex number \(z = a + bi\) as a point \((a,b)\) in the complex plane, where \(a\) is the real part and \(b\) is the imaginary part. Here, \(a = 0\) and \(b = -e\). The modulus \(r\) of \(z\) is \(|z| = \sqrt{a^2 + b^2} = \sqrt{0^2 + (-e)^2} = e\). The argument \(\theta\) is \(\text{arg}(z) = \tan^{-1}\left(\frac{-e}{0}\right)\). Because \(z\) is purely imaginary and negative, \(\theta = -\frac{\pi}{2}\). Hence, the polar form is \(z = e \text{cis} \left(-\frac{\pi}{2}\right)\) or \(e e^{-i\frac{\pi}{2}}\).

Step 2: Use the polar form to express \(\ln z\)

The natural logarithm of a complex number \(z = re^{i\theta}\) is \(\ln z = \ln r + i\theta\). For \(z = e e^{-i\frac{\pi}{2}}\), \(r = e \) and \(\theta = -\frac{\pi}{2}\). So, \(\ln z = \ln e + i(-\frac{\pi}{2})\).

Step 3: Simplify the expression

Using the fact that \(\ln e = 1\), we can substitute to get: \(\ln z = 1 - i\frac{\pi}{2}\). Thus, \(a = 1\) and \(b = -\frac{\pi}{2}\).

Key Concepts

Logarithm of a Complex Number

When dealing with complex numbers, calculating the logarithm can be slightly more nuanced than with real numbers. Let's unpack how you can determine the logarithm of a complex number, specifically in the form of \(z = r \, e^{i\theta}\), where \(r\) is the modulus, and \(\theta\) is the argument.

For a complex number given in polar form, its natural logarithm is calculated as follows:

• Identify the modulus \(r\) of the complex number, which is the distance from the origin to the point \(z\) in the complex plane.
• Determine the argument \(\theta\), which is the angle from the positive x-axis to the line segment representing the complex number.
• The natural logarithm \(\ln(z)\) is expressed as \(\ln(r) + i\theta\).

For example, if \(z = e \, e^{-i\frac{\pi}{2}}\), the logarithm is computed as \(\ln(e) + i\left(-\frac{\pi}{2}\right)\).

Remember that complex logarithms can have multiple values due to the periodic nature of the argument, but typically, we use the principal value for simplicity.

Polar Form of Complex Numbers

The polar form of complex numbers is a powerful way to express numbers as a product of a magnitude (modulus) and a direction (argument).

Each complex number can be represented as a point \( (a, b)\) in the complex plane, where \(a\) is the real part, and \(b\) is the imaginary part. However, that might not always be the most convenient representation, especially for multiplication or division.

Using the polar form, any complex number \(z = a + bi\) is represented as \( r(\cos\theta + i\sin\theta)\), or compactly \( re^{i\theta}\). Here’s how:

• The modulus \(r\) is calculated using \(|z| = \sqrt{a^2 + b^2}\), which gives the 'length' of \(z\) from the origin.
• The argument \(\theta\) is retrieved using \(\tan^{-1}\left(\frac{b}{a}\right)\).
• The polar form helps in simplifying operations such as multiplication and finding powers.

For complex number \(z = -ei\), the modulus is \(e\) and it is purely imaginary, providing a distinct representation in polar form as \(e e^{-i\frac{\pi}{2}}\).

Modulus and Argument

Understanding the modulus and argument of a complex number is key to navigating the complex plane effectively.

The modulus, denoted as \(|z|\), represents the distance from the origin to the complex number \(z\). It is computed using the formula \(|z| = \sqrt{a^2 + b^2}\), where \(a\) is the real part, and \(b\) is the imaginary part.

The argument, referred to as \(\arg(z)\), is the angle formed between the positive x-axis and the line segment joining the origin to the point \(z\). This angle can be found using the arctangent function: \(\arg(z) = \tan^{-1}\left(\frac{b}{a}\right)\).

• The modulus helps in determining the size of the complex number.
• The argument defines the direction or angle the complex number makes.

Combining these components provides the polar representation which can simplify many mathematical operations. For instance, a complex number \( -ei\) has a modulus of \(e\) and an argument of \(-\frac{\pi}{2}\), meaning it lies on the negative imaginary axis.