Problem 82 - Precalculus

Question

Calculate \(\frac{(1+i)^{n}}{(1-i)^{m}},\) where \(n\) and \(m\) are positive integers.

Step-by-Step Solution

Answer

The expression simplifies to \((\sqrt{2})^{n-m}(\cos((n+m)\frac{\pi}{4}) + i\sin((n+m)\frac{\pi}{4}))\).

Step 1: Separate the Complex Numbers

To solve the problem, begin by identifying the complex numbers in the fraction: \((1+i)\) and \((1-i)\). These terms must be treated separately before combining them back together. They both are located in the complex plane.

Step 2: Convert to Polar Form

Convert the complex numbers \((1+i)\) and \((1-i)\) into their polar forms. The modulus (or magnitude) of each is calculated using \(\sqrt{1^2 + 1^2} = \sqrt{2}\). The argument for \(1+i\) is \(\frac{\pi}{4}\) (or 45 degrees), and for \(1-i\) is \(-\frac{\pi}{4}\) (or -45 degrees). Hence, \(1+i = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})\) and \(1-i = \sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))\).

Step 3: Raise to Powers in Polar Form

Use De Moivre's Theorem to raise complex numbers to powers in polar form. For \((1+i)^{n}\), it is \((\sqrt{2})^{n}(\cos(n\frac{\pi}{4}) + i\sin(n\frac{\pi}{4}))\). For \((1-i)^{m}\), it is \((\sqrt{2})^{m}(\cos(-m\frac{\pi}{4}) + i\sin(-m\frac{\pi}{4}))\).

Step 4: Simplify the Entire Expression

Combine the polar expressions: \[\frac{(1+i)^{n}}{(1-i)^{m}} = \frac{(\sqrt{2})^{n}(\cos(n\frac{\pi}{4}) + i\sin(n\frac{\pi}{4}))}{(\sqrt{2})^{m}(\cos(-m\frac{\pi}{4}) + i\sin(-m\frac{\pi}{4}))}\]This simplifies to:\[(\sqrt{2})^{n-m} \left(\frac{\cos(n\frac{\pi}{4}) + i\sin(n\frac{\pi}{4})}{\cos(-m\frac{\pi}{4}) + i\sin(-m\frac{\pi}{4})}\right)\]

Step 5: Final Simplification with Trigonometric Identities

Use the properties of angle addition to simplify the expression further by subtraction of angles. The result is:\[(\sqrt{2})^{n-m} \left( \cos((n+m)\frac{\pi}{4}) + i \sin((n+m)\frac{\pi}{4}) \right)\]Express this back in rectangular form or leave it in polar/exponential form as desired. The final value can dynamically change with different \(n\) and \(m\).

Key Concepts

Polar Form

Polar form is a way to express complex numbers by using a magnitude and an angle. This is useful when dealing with multiplication and division. A complex number in rectangular form looks like this:

• Rectangular form: a + bi, where 'a' and 'b' are real numbers, and 'i' is the imaginary unit (\( i^2 = -1 \)).

To convert a complex number from rectangular form to polar form:

• Calculate the modulus (magnitude): \[ r = \sqrt{a^2 + b^2} \]
• Determine the argument (angle), \( \theta \), using the inverse tangent function: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \]
• The polar form then becomes: \[ r(\cos \theta + i\sin \theta) \]

For example, the complex number \((1+i)\) has a magnitude, \(\sqrt{2}\), and an angle, \(\frac{\pi}{4}\), leading to the polar form: \[ \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) \].This format simplifies multiplication and division, allowing us to easily manipulate and simplify expressions involving complex numbers.

De Moivre's Theorem

De Moivre's Theorem facilitates raising complex numbers in polar form to a power. It is especially useful for calculating powers and roots of complex numbers, which can be complex and tedious otherwise. The theorem states:

• For a complex number in polar form: \[ r(\cos \theta + i\sin \theta) \]
• Raising it to the power n: \[ r^n(\cos(n\theta) + i\sin(n\theta)) \]

This theorem allows you to multiply the angle \( \theta \) by the exponent \( n \), and raise the modulus \( r \) to the power \( n \).
This powerful tool transforms lengthy calculations into straightforward operations. In our example with \((1+i)^n\), the modulus \(\sqrt{2}\) becomes \((\sqrt{2})^n\) and the angle \(\frac{\pi}{4}\) is multiplied by \( n \). Thus, making complex calculations simpler and more efficient.

Trigonometric Identities

Understanding trigonometric identities is key to simplifying expressions with complex numbers. They allow us to transform and simplify expressions effectively. Here are a few identities that are particularly useful:

• Addition and Subtraction: \[ \cos(a \pm b) = \cos a \cos b \mp \sin a \sin b \]
• Sin Relations: \[ \sin(a \pm b) = \sin a \cos b \pm \cos a \sin b \]

In the exercise, after using De Moivre's Theorem, we ended up dividing polar forms with angles \( n\frac{\pi}{4} \) and \(-m\frac{\pi}{4} \).
Using trigonometric identities allows us to further reduce the expression:

• In the form \( \cos((n+m)\frac{\pi}{4}) + i \sin((n+m)\frac{\pi}{4}) \), which is now cleaner and more manageable.

These identities are essential for transforming complex operations into simpler forms, making them all the more understandable and allowing direct application to varying values of \( n \) and \( m \).