Problem 19 - Advanced Engineering Mathematics

Question

In Problems 17-20, write the given number in the form \(a+i b\). $$ \frac{(1-i)^{10}}{(1+i)^{3}} $$

Step-by-Step Solution

Answer

The expression simplifies to \(4\sqrt{2} + 4\sqrt{2}i\).

Step 1: Interpret the Expression

We must express \( \frac{(1-i)^{10}}{(1+i)^{3}} \) in the form \( a + ib \). This involves simplifying the expression and extracting the real and imaginary parts.

Step 2: Simplify the Numerator

First, we will simplify \((1-i)^{10}\) using De Moivre's theorem. Express \(1-i\) in polar form as \( \sqrt{2} \text{cis}(-\frac{\pi}{4}) \), so \((1-i)^{10} = (\sqrt{2})^{10} \text{cis}(-\frac{10\pi}{4}) = 32 \text{cis}(-\frac{5\pi}{2}) \).

Step 3: Simplify the Denominator

Now simplify \((1+i)^{3}\). Express \(1+i\) in polar form as \( \sqrt{2} \text{cis}(\frac{\pi}{4}) \), so \((1+i)^{3} = (\sqrt{2})^{3} \text{cis}(\frac{3\pi}{4}) = 2\sqrt{2} \text{cis}(\frac{3\pi}{4}) \).

Step 4: Combine Numerator and Denominator

Substitute the simplified forms into the original expression: \( \frac{32 \text{cis}(-\frac{5\pi}{2})}{2\sqrt{2} \text{cis}(\frac{3\pi}{4})} = \frac{32}{2\sqrt{2}} \text{cis}(-\frac{5\pi}{2} - \frac{3\pi}{4}) = 8 \text{cis}(-\frac{17\pi}{4}) \).

Step 5: Convert Back to Rectangular Form

The angle \(-\frac{17\pi}{4}\) is equivalent to \(\frac{\pi}{4}\) (since -17\pi/4 + 4\pi = \pi/4), so \(8 \text{cis}(\frac{\pi}{4}) = 8(\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4})) = 8\left(\frac{\sqrt{2}}{2}\right) + 8i\left(\frac{\sqrt{2}}{2}\right) = 4\sqrt{2} + 4\sqrt{2}i \).

Step 6: Conclusion

Therefore, the expression \(\frac{(1-i)^{10}}{(1+i)^{3}}\) simplifies to \(4\sqrt{2} + 4\sqrt{2}i\), which is the form \(a + ib\) with \(a = 4\sqrt{2}\) and \(b = 4\sqrt{2}\).

Key Concepts

De Moivre's Theorem

De Moivre's theorem is a powerful tool in complex number mathematics, especially useful when working with powers of complex numbers in polar form. It provides a simple way to compute powers and roots of complex numbers. According to this theorem, if a complex number is given in its polar form as \[z = r(\cos \theta + i \sin \theta) = r \text{cis} \theta,\]then \[z^n = r^n (\cos(n\theta) + i \sin(n\theta)) = r^n \text{cis}(n\theta).\]This theorem simplifies what would otherwise be a complex and lengthy multiplication process. It breaks down the problem into smaller, manageable parts, allowing you to focus on multiplying the modulus \(r\) and multiplying the argument \(\theta\) by the power \(n\).
By employing De Moivre's theorem, you can straightforwardly compute powers of a complex number once expressed in polar form, as showcased in the original solution.

Polar Form

The polar form is a representation of complex numbers that extends the idea of the modulus and argument. In contrast to the regular rectangular format \(a + ib\), polar form is represented by the modulus \(r\) and the argument \(\theta\), which describes a point on the complex plane.
The polar form of a complex number \(z\) is \[z = r \text{cis} \theta,\]where \(r\) is the modulus given by the equation \(r = \sqrt{a^2 + b^2}\) and \(\theta\) is the argument given by \(\theta = \tan^{-1}(\frac{b}{a})\).
This format is incredibly useful for multiplication and division, as it simplifies these operations to scaling and rotating in the complex plane. You convert to polar form by determining the distance and angle from the origin to the point \(a + ib\), transforming intricate operations into more manageable tasks.

Rectangular Form

The rectangular form of a complex number is probably the most familiar format. It expresses complex numbers as the sum of a real part and an imaginary part, denoted by \(a + ib\). Here, \(a\) is the real component, and \(b\) is the imaginary coefficient.
This form aligns well with how we perform addition and subtraction with complex numbers. However, it can become cumbersome when involving operations like multiplication, division, or raising to powers, which is where polar form shines.
For any operation returning a complex number, it's often necessary to convert from polar back to rectangular form. This conversion involves applying trigonometric functions on the polar expression, such as \[z = r \text{cis} \theta = r(\cos \theta + i \sin \theta) = r \cos \theta + i r \sin \theta.\]Thus, the original problem solution ends with converting back to \(a + ib\) by determining the values of \(a\) and \(b\) from the modulus and argument.