Problem 44 - Advanced Engineering Mathematics

Question

Find the principal value of the given quantity. Express answers in the form \(a+i b\). \((1-i)^{2 i}\)

Step-by-Step Solution

Answer

The principal value is \(e^{\frac{\pi}{2}} \cos(\ln 2) + i e^{\frac{\pi}{2}} \sin(\ln 2)\).

Step 1: Express the Complex Number in Polar Form

The complex number \((1-i)\) can be expressed in polar form. First, we calculate the modulus: \(|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\). Next, we find the argument: \(\tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}\). Thus, the polar form is \(\sqrt{2} \text{cis} \left(-\frac{\pi}{4}\right)\).

Step 2: Apply the Exponentiation

Using De Moivre's theorem, \((\sqrt{2} \text{cis}(-\frac{\pi}{4}))^{2i}\) becomes \((\sqrt{2})^{2i} \text{cis}(2i \cdot -\frac{\pi}{4})\). This simplifies to \((2^{i}) \cdot \text{cis}(-\frac{\pi}{2}i)\).

Step 3: Simplify with Euler's Formula

The expression \(2^{i}\) can be represented using Euler's formula as \(e^{i \ln 2} = \cos(\ln 2) + i \sin(\ln 2)\). Meanwhile, \(\text{cis}(-\frac{\pi}{2}i) = e^{-\frac{\pi}{2}i^2} = e^{\frac{\pi}{2}}\).

Step 4: Combine the Results

Now combine the terms from Step 3: \( (\cos(\ln 2) + i \sin(\ln 2)) \cdot e^{\frac{\pi}{2}}\). This gives the expression in form \(a+ib\): \(a = e^{\frac{\pi}{2}} \cos(\ln 2)\) and \(b = e^{\frac{\pi}{2}} \sin(\ln 2)\).

Step 5: Final Expression

Therefore, the principal value of \((1-i)^{2i}\) is \(e^{\frac{\pi}{2}} \cos(\ln 2) + i e^{\frac{\pi}{2}} \sin(\ln 2)\).

Key Concepts

Polar Form of Complex Numbers

When working with complex numbers, it's often more convenient to express them in polar form than in the standard form. The polar form represents a complex number in terms of its magnitude (or modulus) and its angle (or argument) relative to the positive x-axis.

A complex number like \(a + bi\) can be converted to polar form as follows:

• Calculate the modulus: \(|z| = \sqrt{a^2 + b^2}\).
• Determine the argument: \(\text{arg}(z) = \tan^{-1}\left(\frac{b}{a}\right)\).

Using these formulas, the complex number \(1-i\) can be calculated to have:

• Modulus: \(\sqrt{2}\)
• Argument: \(-\frac{\pi}{4}\)

Hence, its polar form is denoted as \(\sqrt{2} \text{cis} \left(-\frac{\pi}{4}\right)\), where \(\text{cis} \theta = \cos \theta + i \sin \theta\).

This polar representation is particularly useful in operations like multiplication, division, and, as we'll see later, exponentiation.

De Moivre's Theorem

De Moivre's Theorem is a powerful tool in complex number analysis, especially useful for finding powers and roots of complex numbers. The theorem states that for a complex number in polar form, \((r \text{cis} \theta)^n = r^n \text{cis}(n\theta)\).

In our problem, we used De Moivre's Theorem to exponentiate the polar form of \(1-i\) to the power of \(2i\). That involved:

• Raising the modulus to the power \(2i\): \((\sqrt{2})^{2i}\).
• Multiplying the argument by \(2i\): \(\text{cis}(-\frac{\pi}{2}i)\).

This usage simplifies the calculation dramatically because the polar form directly utilizes angles and magnitudes to handle complex operations with ease, further laying the groundwork to incorporate Euler's Formula.

Euler's Formula

Euler's Formula bridges complex numbers and exponential functions in an elegant way. It states that for any real number \(x\), the complex exponential \(e^{ix}\) equals \(\cos(x) + i\sin(x)\). This connection is incredibly useful when dealing with complex exponentiation and logarithmic transformations.

In the given problem, we employed Euler's Formula to transform complex exponentiations:

• \(2^i\) is converted into \(e^{i \ln 2}\).
• This becomes \(\cos(\ln 2) + i \sin(\ln 2)\).

We also used the formula for simplifying \(\text{cis}(-\frac{\pi}{2}i)\) into an exponential form. This translates to \(e^{\frac{\pi}{2}}\), ultimately converting complex trigonometric expressions into an easily manageable exponential form.

By combining the outputs from Euler's Formula with results from other steps, we derived the final expression in standard \(a + ib\) format, effortlessly converting between different mathematical perspectives.