De Moivre’s Theorem: Unlocking the Power of Complex Numbers
Welcome to one of the most powerful and elegant theorems in Further Mathematics! De Moivre's Theorem (often abbreviated D.M.T.) is the essential bridge that connects complex numbers (in their polar form) with trigonometric functions (like sine and cosine).
In this chapter, you will learn how to use this theorem to raise complex numbers to any integer power, find roots easily, and derive complicated trigonometric identities, which is crucial for tackling challenging integration problems.
Why is this important? Before D.M.T., finding \( (3 + 4i)^{10} \) would be a nightmarish task involving multiplying brackets ten times. D.M.T. turns it into a simple multiplication problem!
1. The Foundation: Polar Form Prerequisite
De Moivre's Theorem only works directly when the complex number \( z \) is written in its polar form, \( z = r(\cos \theta + i \sin \theta) \).
Recall that:
- Modulus (\( r \)): The distance from the origin on the Argand diagram. \( r = |z| \).
- Argument (\( \theta \)): The angle (in radians) measured anticlockwise from the positive real axis. \( \theta = \arg(z) \).
Memory Aid: Sometimes \(\cos \theta + i \sin \theta\) is shortened to \(\text{cis}\,\theta\). While you shouldn't use \(\text{cis}\,\theta\) in formal answers, it's a great way to remember the structure!
The Core Theorem (De Moivre's Theorem for Integral \(n\))
If \( z = r(\cos \theta + i \sin \theta) \) and \( n \) is an integer (positive or negative), then:
$$ z^n = r^n (\cos(n\theta) + i \sin(n\theta)) $$
What does this mean?
- To raise \( z \) to the power \( n \), you raise the modulus \( r \) to the power \( n \).
- You simply multiply the argument \( \theta \) by \( n \).
Analogy: Imagine a complex number as a vector that has a length (\( r \)) and a direction (\( \theta \)). Raising it to the power \( n \) is like making the vector \( n \) times longer (in terms of scaling its area/volume) and rotating it \( n \) times further.
Quick Review: Key Takeaway
D.M.T. turns the difficult process of raising complex numbers to powers into the easy process of multiplying angles.
2. Applications in Trigonometry and Integration
One of the most important applications of D.M.T. is deriving trigonometric identities, especially those involving powers of trig functions (like \(\cos^5 \theta\)) and multiple angle formulas (like \(\cos 4\theta\)).
Relating Powers of \(z\) to Trig Functions
Let's consider a complex number \( z \) where the modulus \( r=1 \). So, \( z = \cos \theta + i \sin \theta \).
By D.M.T., its reciprocal \( \frac{1}{z} \) (which is \( z^{-1} \)) is:
$$ \frac{1}{z} = \cos(-\theta) + i \sin(-\theta) = \cos \theta - i \sin \theta $$
Adding and subtracting these two expressions gives us the essential identities for this section:
$$ z + \frac{1}{z} = (\cos \theta + i \sin \theta) + (\cos \theta - i \sin \theta) = 2 \cos \theta $$
$$ z - \frac{1}{z} = (\cos \theta + i \sin \theta) - (\cos \theta - i \sin \theta) = 2i \sin \theta $$
More generally, using D.M.T. again:
$$ z^n + \frac{1}{z^n} = 2 \cos(n\theta) $$
$$ z^n - \frac{1}{z^n} = 2i \sin(n\theta) $$
A. Expressing Powers of \(\cos \theta\) and \(\sin \theta\) as Multiple Angles
This technique is essential for evaluating integrals where the integrand is a high power of a trig function, e.g., \( \int \cos^4 \theta \, d\theta \).
Step-by-Step Process for \(\cos^n \theta\)
- Start with the identity: \( 2 \cos \theta = z + \frac{1}{z} \).
- Raise both sides to the power \( n \): \( (2 \cos \theta)^n = \left(z + \frac{1}{z}\right)^n \).
- Expand the right side using the Binomial Theorem.
- Group the terms in conjugate pairs: \( \left(z^k + \frac{1}{z^k}\right) \).
- Substitute back using the D.M.T. identity: \( z^k + \frac{1}{z^k} = 2 \cos(k\theta) \).
- Isolate \(\cos^n \theta\).
Example for Integration: If you need to integrate \(\cos^4 \theta\), you would derive the formula:
$$ \cos^4 \theta = \frac{1}{8}(\cos 4\theta + 4 \cos 2\theta + 3) $$
Integrating the right side is now trivial!
Common Mistake to Avoid: When dealing with \(\sin^n \theta\), remember the factor of \( i \)! You use \( (2i \sin \theta)^n = \left(z - \frac{1}{z}\right)^n \). The powers of \( i \) (which cycle \( i, -1, -i, 1 \)) must be correctly handled.
B. Expressing Multiple Angle Functions in terms of \(\tan \theta\)
The syllabus requires you to be able to express functions like \(\tan 5\theta\) in terms of powers of \(\tan \theta\).
1. Start by finding the expression for \(\cos(n\theta) + i \sin(n\theta)\) using two methods:
- Method 1 (D.M.T.): \( \cos(n\theta) + i \sin(n\theta) = (\cos \theta + i \sin \theta)^n \).
- Method 2 (Binomial Theorem): Expand \( (\cos \theta + i \sin \theta)^n \).
2. Equate the expanded Binomial expression to the D.M.T. result. By comparing the real parts, you get \(\cos n\theta\). By comparing the imaginary parts, you get \(\sin n\theta\).
3. To find \(\tan n\theta\), use the identity: \( \tan n\theta = \frac{\sin n\theta}{\cos n\theta} \).
4. Divide the numerator and denominator by the highest power of \(\cos \theta\) (which is \(\cos^n \theta\)) to convert all terms into powers of \(\tan \theta\).
Did you know? These multiple angle formulae (especially for \(\cos n\theta\)) are closely related to Chebyshev Polynomials, showing how complex numbers link to advanced algebra!
Quick Review: Key Takeaway
The identities \( 2 \cos \theta = z + \frac{1}{z} \) and \( 2i \sin \theta = z - \frac{1}{z} \) are fundamental tools for manipulating trigonometric expressions for use in calculus.
3. Finding the \(n\)th Roots of Complex Numbers
De Moivre's Theorem is equally useful for solving equations of the form \( z^n = w \), where \( w = a + ib \) is a complex number.
The process is to find \( w^{1/n} \).
The Need for Multiple Arguments
When finding roots, we must account for the fact that the argument of a complex number is periodic. Since a rotation of \( 2\pi \) returns you to the same position, \( w \) can be written as:
$$ w = r(\cos(\theta + 2k\pi) + i \sin(\theta + 2k\pi)) $$
where \( k \) is any integer.
If we apply D.M.T. to find the \( n \)th root (i.e., power \( \frac{1}{n} \)), we get the general formula for the roots \( z_k \):
$$ z_k = r^{1/n} \left(\cos \left(\frac{\theta + 2k\pi}{n}\right) + i \sin \left(\frac{\theta + 2k\pi}{n}\right)\right) $$
Solving \( z^n = a + ib \)
Step-by-Step Root Finding
- Convert \( w = a + ib \) to Polar Form: Find the modulus \( r \) and the principal argument \( \theta \) (where \( -\pi < \theta \le \pi \)).
- Apply the General Formula: Write the root formula incorporating the \( 2k\pi \).
- Find the \( n \) Distinct Roots: Use integer values of \( k = 0, 1, 2, \dots, n-1 \). These \( n \) values will give you the \( n \) distinct roots. (Once \( k=n \), the angles repeat).
- Express in Required Form: The question might ask for the answer in Cartesian form (\( x + iy \)) or surd form (if the angles are related to exact trig values).
Roots of Unity
The simplest and most beautiful case is finding the solutions to \( z^n = 1 \). These are the \( n \)th roots of unity.
Since \( 1 = 1(\cos 0 + i \sin 0) \), the roots are:
$$ z_k = 1^{1/n} \left(\cos \left(\frac{2k\pi}{n}\right) + i \sin \left(\frac{2k\pi}{n}\right)\right) $$
for \( k = 0, 1, \dots, n-1 \).
Geometric Interpretation: These roots are equally spaced around the unit circle on the Argand diagram, forming the vertices of a regular \( n \)-sided polygon. One vertex is always at \( (1, 0) \) (when \( k=0 \)).
Quick Review: Key Takeaway
When finding the \( n \)th roots, remember to include the periodicity \( (+ 2k\pi) \) in the argument. You must find exactly \( n \) distinct roots by varying \( k \).
4. Euler's Formula and the Exponential Form
While D.M.T. is powerful, the ultimate simplification comes from relating the polar form to the exponential function. You are expected to know and use this identity without proof.
Euler’s Formula
$$ e^{i\theta} = \cos \theta + i \sin \theta $$
This is the exponential form of a complex number with modulus 1. If the modulus is \( r \), the complex number is written as:
$$ z = r e^{i\theta} $$
Why is this form so useful?
1. It makes D.M.T. trivial: If \( z = r e^{i\theta} \), then \( z^n = (r e^{i\theta})^n = r^n e^{in\theta} \). This immediately converts back to D.M.T. in polar form: \( r^n (\cos n\theta + i \sin n\theta) \).
2. Multiplication/Division: Multiplying complex numbers means adding the arguments, and dividing means subtracting the arguments—just like exponents!
$$ z_1 z_2 = (r_1 e^{i\theta_1}) (r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_1 + \theta_2)} $$
Application: Calculations involving Surd Form
The exponential form is often used when dealing with exact values that require manipulating angles that aren't 'standard' but can be composed of standard angles. The syllabus specifically mentions expressions like \(\cos \frac{5\pi}{12}\) in surd form.
To find \(\cos \frac{5\pi}{12}\):
- Recognise that \( \frac{5\pi}{12} \) can be split into standard angles: \( \frac{5\pi}{12} = \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{\pi}{6} + \frac{\pi}{4} \).
- Use the exponential form: Calculate \( e^{i\pi/6} \cdot e^{i\pi/4} \).
- Convert to Cartesian form and multiply:
$$ e^{i\pi/6} e^{i\pi/4} = \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) $$
$$ = \left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i\right) $$ - Expand and equate the real part: The real part of the product must equal \( \cos(\frac{\pi}{6} + \frac{\pi}{4}) = \cos \frac{5\pi}{12} \).
Don't worry if this seems tricky at first—it’s just a systematic way of applying addition formulae using the complex number framework!
Quick Review: Key Takeaway
Euler’s formula \( e^{i\theta} = \cos \theta + i \sin \theta \) is the most compact way to express and calculate powers and products of complex numbers, making D.M.T. simple exponentiation.
5. Chapter Summary and Formula Review
De Moivre's Theorem is a cornerstone of FP2. Mastering it means mastering these key relationships:
The Big Three Identities
- D.M.T. (Integral \( n \)):
$$ (\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta) $$ - Euler's Formula (Exponential Form):
$$ e^{i\theta} = \cos \theta + i \sin \theta $$ - Root Formula (The General Solution for \( z^n = w \)):
$$ z_k = r^{1/n} e^{i(\frac{\theta + 2k\pi}{n})}, \quad k = 0, 1, \dots, n-1 $$
Tools for Trigonometric Manipulations
These identities are crucial for converting powers of trigonometric functions into multiple angles (for integration):
- $$ z^n + \frac{1}{z^n} = 2 \cos(n\theta) $$
- $$ z^n - \frac{1}{z^n} = 2i \sin(n\theta) $$
Keep practising the conversions between Cartesian, Polar, and Exponential forms. This is the key to unlocking the power of De Moivre's Theorem!