Problem 25 - Advanced Engineering Mathematics

Question

Express \(\ln z\) in the form \(a+i b\). \(z=-2+2 i\)

Step-by-Step Solution

Answer

\(\ln z = \frac{3}{2} \ln 2 + i \frac{3\pi}{4}\).

Step 1: Convert to Polar Form

We start by expressing the complex number \(z = -2 + 2i\) in its polar form. The polar form is given by \(r(\cos \theta + i \sin \theta)\) where \(r\) is the modulus and \(\theta\) is the argument. First, calculate the modulus \(r\): \[ r = \sqrt{(-2)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}. \] Next, find the argument \(\theta\) using the tangent function: \[ \theta = \text{atan2}(2, -2) = \frac{3\pi}{4}. \] So, the polar form is \(2\sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})\).

Step 2: Use Logarithm Properties

The natural logarithm of a complex number \(z\) in polar form \(re^{i\theta}\) is given by \(\ln z = \ln r + i\theta\). Using the polar form from Step 1: \[ \ln z = \ln(2\sqrt{2}) + i \frac{3\pi}{4}. \] The modulus \(r = 2\sqrt{2}\) can be simplified as: \[ \ln(2\sqrt{2}) = \ln 2 + \ln \sqrt{2} = \ln 2 + \frac{1}{2} \ln 2 = \frac{3}{2} \ln 2. \] Now substituting back, we have: \[ \ln z = \frac{3}{2} \ln 2 + i \frac{3\pi}{4}. \]

Step 3: Express in the Form a + ib

We translate the expression obtained into the form \(a + ib\), where \(a\) is the real part and \(b\) is the imaginary part. From Step 2, we found: \[ \ln z = \frac{3}{2} \ln 2 + i \frac{3\pi}{4}. \] Thus, \(a = \frac{3}{2} \ln 2\) and \(b = \frac{3\pi}{4}\).

Key Concepts

Polar Form

When dealing with complex numbers, transforming a complex number from its standard form to its polar form can greatly simplify many operations, such as multiplication and determining powers or roots. The polar form expresses a complex number using a combination of its modulus and argument.

To explain in detail, consider a complex number, which can be expressed as \(z = a + bi\). Here, `a` is the real part and `b` is the imaginary part. The polar form represents this as \(r(\cos \theta + i \sin \theta)\), or more compactly, \(z = re^{i\theta}\).

Some useful elements:

• The modulus \(r\) is the distance from the origin to the point \((a, b)\) in the complex plane. It measures the "size" of the complex number and is calculated as \(r = \sqrt{a^2 + b^2}\).
• The argument \(\theta\) is the angle made with the positive real axis. Calculating \(\theta\) involves using the inverse tangent function, commonly noted as \(\theta = \text{atan2}(b, a)\).

In our original exercise, the complex number \(-2 + 2i\) was converted to its polar form by first finding \(r = 2\sqrt{2}\) and \(\theta = \frac{3\pi}{4}\). Knowing how to perform these calculations is critical when working with complex numbers in various applications.

Complex Logarithm

The logarithm of a complex number can be a bit more challenging than real numbers because it involves both magnitude and direction. One major trick to simplify this is using the polar form.

Once a complex number \(z = re^{i\theta}\) is expressed in polar form, finding its natural logarithm, or \(\ln z\), becomes more straightforward. The formula: \[\ln z = \ln r + i\theta\]

This approach separates the modulus \(r\) and argument \(\theta\), allowing you to calculate each part independently.

In our worked example, \(z = -2 + 2i\) was converted into \(r = 2\sqrt{2}\) and \(\theta = \frac{3\pi}{4}\). This led to:

• \(\ln r = \ln(2\sqrt{2}) = \frac{3}{2} \ln 2\)
• The imaginary component directly follows as \(\theta = \frac{3\pi}{4}\)

The result beautifully combines into \(\ln z = \frac{3}{2} \ln 2 + i\frac{3\pi}{4}\), splitting into its real and imaginary parts, suitable for many applications in mathematics and engineering.

Modulus and Argument

Understanding the modulus and argument is key in dealing with complex numbers, especially when operations require transformations, like converting to polar form, as previously discussed. The modulus, or absolute value, provides the "size" or "magnitude" of the complex number.

Given \(z = a + bi\), the modulus is computed as \(\sqrt{a^2 + b^2}\), offering a geometric interpretation as it represents the length of the vector in the complex plane.

The argument, on the other hand, provides the orientation of this vector, delivering the angle with respect to the positive real axis. Calculating this angle is simple with \( \theta = \text{atan2}(b, a)\). This function, \(\text{atan2}\), handles the appropriate quadrant, overcoming limitations of standard \(\tan^{-1}\), especially when working across different quadrants.

For a given complex number like \(-2 + 2i\), the modulus becomes \(2\sqrt{2}\), and the argument is \(\frac{3\pi}{4}\). These values enable the transition into polar form, unveiling new methods for complex arithmetic and algebra, making modulus and argument elementary yet powerful concepts in understanding complex numbers.