Problem 10 - Complex Analysis for Mathematics and Engineering

Question

Let \(m\) and \(n\) be positive integers that have no common factor. Show that there are \(n\) distinct solutions to \(w^{n}=z^{m}\) and that they are given by \(w_{k}=r^{m}\left(\cos \frac{m(\theta+2 \pi k)}{n}+i \sin \frac{m(\theta+2 \pi k)}{n}\right)\) for \(k=0,1, \ldots, n-1 .\)

Step-by-Step Solution

Answer

There are \(n\) distinct solutions given by \(w_k = r^{m/n} \left( \cos \frac{m(\theta + 2\pi k)}{n} + i \sin \frac{m(\theta + 2\pi k)}{n} \right)\) for \(k = 0, 1, \ldots, n-1\).

Step 1: Understand the Problem

We need to find the solutions to the equation \(w^n = z^m\) where \(w\) and \(z\) are complex numbers, \(m\) and \(n\) are positive integers with no common factor. We also have to show these solutions are distinct and given in the specified form.

Step 2: Polar Representation of Complex Numbers

We represent a complex number \(z\) in polar form, \(z = r \left( \cos \theta + i \sin \theta \right)\), where \(r\) is the magnitude and \(\theta\) is the argument of the complex number.

Step 3: Apply De Moivre's Theorem

By De Moivre's Theorem, a complex number raised to a power can be written as: \(z^m = r^m \left( \cos(m \theta) + i \sin(m \theta) \right)\). This gives us \(w^n = z^m = r^m \left( \cos(m \theta) + i \sin(m \theta) \right)\).

Step 4: Find General Solution for \(w\)

To solve for \(w\), we take the \(n^{th}\) root of both sides: \(w = r^{m/n} \left( \cos\left(\frac{m \theta + 2 \pi k}{n}\right) + i \sin\left(\frac{m \theta + 2 \pi k}{n}\right)\right)\) for integer \(k\) since we consider roots in terms of full rotations around the unit circle.

Step 5: Show Distinct Solutions

Because \(m\) and \(n\) have no common factor, the angles \(\frac{m(\theta + 2\pi k)}{n}\) for \(k = 0, 1, \ldots, n-1\) correspond to distinct angles due to their alignment covering all distinct roots on the circle exactly once.

Step 6: Conclude with the Set of Solutions

Thus, the distinct solutions to \(w^n = z^m\) are: \(w_k = r^{m/n} \left( \cos \frac{m(\theta + 2\pi k)}{n} + i \sin \frac{m(\theta + 2\pi k)}{n} \right)\) for each \(k = 0, 1, \ldots, n-1\).

Key Concepts

De Moivre's Theorem

De Moivre's Theorem is a very useful tool when dealing with powers and roots of complex numbers.
It states that for any complex number in polar form, and any integer exponent, we can easily calculate the power:

• If we have a complex number represented as \(z = r(\cos \theta + i \sin \theta)\), then \(z^n = r^n(\cos(n\theta) + i \sin(n\theta))\).

This is incredibly helpful because it transforms the tricky multiplication of complex numbers into simple operations with trigonometric components.
In our problem, by using De Moivre's Theorem, we are able to express \(z^m\) and find solutions for \(w^n = z^m\) efficiently.
After applying the theorem, obtaining the nth roots is straightforward:

• We take the nth root of the magnitude \(r^m\), resulting in \(r^{m/n}\).


• The angle becomes \(\frac{m \theta}{n} + \frac{2\pi k}{n}\), representing rotations around the circle.

Complex Numbers

Complex numbers are extensions of real numbers. A complex number can be written as \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit satisfying \(i^2 = -1\).
They can also be represented in polar form:

• \(z = r(\cos \theta + i \sin \theta)\)

where:

• \(r\) is the modulus (or magnitude) \(\sqrt{a^2 + b^2}\)
• \(\theta\) is the argument (or angle) of the complex number

Using polar coordinates simplifies the manipulation of complex numbers.
It allows us to conveniently apply De Moivre's Theorem, which is critical for solving equations like \(w^n = z^m\).
This exercise specifically benefits from turning complex numbers to polar form to identify and calculate all distinct roots by examining their geometric representation on the unit circle.

Roots of Unity

Roots of unity are special complex numbers.
They are solutions to equations of the form \(x^n = 1\), perfectly spaced on the unit circle in the complex plane.
The general form for these roots is:

• \(e^{2 \pi i k / n}\) for integer \(k = 0, 1, \ldots, n-1\)

These roots are fundamental in many areas of complex analysis and number theory.
In our problem, the concept of roots of unity helps. By finding the distinct \(n\)-th roots of a complex number \(z^m\), which lie uniformly on the circle.
Since \(m\) and \(n\) have no common factor, \(k\) in the angle \(\frac{m \theta + 2\pi k}{n}\) ensures each solution is a new point.
These angles correspond to distinct roots around the circle, demonstrating how the original equation yields \(n\) distinct solutions.