🚀 Further Complex Numbers: Mastering Powers and Roots (Unit FP2)
Hello future mathematician! Welcome to the world of Further Complex Numbers. In FP1, you learned how to plot, add, multiply, and divide complex numbers. Here in FP2, we are taking it up several notches. You are about to unlock the mathematical magic that allows us to find high powers and specific roots of complex numbers almost instantly!
Why is this chapter important? It provides powerful tools, specifically De Moivre's Theorem and the Exponential Form, which are essential for solving polynomial equations and linking complex numbers to advanced trigonometry.
1. Quick Review: Polar Form (The Essential Prerequisite)
To use the big theorems in this chapter, you must be comfortable converting between Cartesian form (\(z = x + iy\)) and Polar form (\(z = r(\cos\theta + i\sin\theta)\)).
- Modulus (\(r\)): The distance from the origin. \(r = |z| = \sqrt{x^2 + y^2}\).
- Argument (\(\theta\)): The angle measured anti-clockwise from the positive real axis. Remember to adjust the angle based on the quadrant!
- Principal Argument: We usually use the argument such that \(-\pi < \theta \le \pi\). This makes communication much clearer.
Quick Review Box: If you struggle with finding \(\theta\) in the correct quadrant, stop and practice this now. A mistake in \(\theta\) will ruin all subsequent calculations!
2. De Moivre's Theorem: The Power Shortcut
Imagine needing to calculate \((1 + i)^{20}\). Multiplying this out 20 times would take forever! De Moivre's Theorem provides an elegant solution.
2.1. The Theorem for Integers
If \(z = r(\cos\theta + i\sin\theta)\) and \(n\) is any integer (positive or negative), then:
$$ (r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta)) $$What does this mean? When raising a complex number to the power \(n\):
- Raise the modulus, \(r\), to the power \(n\).
- Multiply the argument, \(\theta\), by the power \(n\).
Analogy: Think of De Moivre as the ultimate mathematical cruise control. Instead of taking many small steps (multiplication), you just set the destination (multiply the angle by the power).
2.2. Step-by-Step Example (Positive Power)
Calculate \((\sqrt{3} + i)^6\).
- Convert to Polar Form:
\(r = |\sqrt{3} + i| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2\).
\(\tan\alpha = 1/\sqrt{3} \implies \alpha = \pi/6\). Since it's in the first quadrant, \(\theta = \pi/6\).
So, \(\sqrt{3} + i = 2(\cos(\pi/6) + i\sin(\pi/6))\). - Apply De Moivre (with \(n=6\)):
\((\sqrt{3} + i)^6 = [2(\cos(\pi/6) + i\sin(\pi/6))]^6\)
\(= 2^6 (\cos(6 \cdot \pi/6) + i\sin(6 \cdot \pi/6))\) - Simplify:
\(= 64 (\cos(\pi) + i\sin(\pi))\)
Since \(\cos(\pi) = -1\) and \(\sin(\pi) = 0\), the answer is:
\(= 64(-1 + i(0)) = \mathbf{-64}\).
Did you know? De Moivre's Theorem is actually a special case of Euler's Formula (which we'll look at next), but it was discovered decades earlier!
Key Takeaway: De Moivre's Theorem transforms power calculation from repeated multiplication into simple angle scaling.
3. The Exponential Form (Euler's Identity)
The exponential form is arguably the most elegant way to write a complex number. It is derived from Euler's Identity.
3.1. Defining the Exponential Form
Euler's Identity states that for any real angle \(\theta\):
$$ \cos\theta + i\sin\theta = \text{e}^{i\theta} $$Therefore, the exponential form of a complex number \(z\) is:
$$ z = r\text{e}^{i\theta} $$where \(r\) is the modulus and \(\theta\) is the argument (in radians).
Why use this?
1. It makes De Moivre look even tidier: \((r\text{e}^{i\theta})^n = r^n\text{e}^{in\theta}\).
2. It simplifies multiplication and division using the rules of indices:
Multiplication: \(z_1 z_2 = (r_1 \text{e}^{i\theta_1})(r_2 \text{e}^{i\theta_2}) = r_1 r_2 \text{e}^{i(\theta_1 + \theta_2)}\)
Division: \(z_1 / z_2 = (r_1 \text{e}^{i\theta_1}) / (r_2 \text{e}^{i\theta_2}) = (r_1 / r_2) \text{e}^{i(\theta_1 - \theta_2)}\)
Mnemonic: Remember that "E" is for Easy! Exponential form makes algebraic manipulation of complex numbers much easier.
Important Note: When using the exponential form for FP2 problems, the argument \(\theta\) must always be in radians.
Key Takeaway: The exponential form \(r\text{e}^{i\theta}\) is a condensed notation that follows standard index laws, making calculations very efficient.
4. Finding the \(n\)-th Roots of a Complex Number
This is the most crucial and often most challenging application of De Moivre's Theorem in FP2. We are solving equations of the form \(z^n = w\), where \(w\) is a known complex number.
The Fundamental Theorem of Algebra guarantees that an equation \(z^n = w\) must have exactly \(n\) solutions (or roots).
4.1. The Critical Concept: Periodic Argument
The key insight is remembering that angles repeat every \(2\pi\) radians (\(360^\circ\)).
For any complex number \(w = R(\cos\Phi + i\sin\Phi)\), we can also write it as:
where \(k\) is any integer (\(k = 0, \pm 1, \pm 2, \dots\)). This is called the general argument.
4.2. The General Formula for Roots
To find the \(n\) roots of \(w = R\text{e}^{i\Phi}\), we solve \(z^n = w\). Using the exponential form with the general argument:
$$ z^n = R\text{e}^{i(\Phi + 2k\pi)} $$Taking the \(n\)-th root of both sides:
$$ z_k = \sqrt[n]{R} \text{e}^{i\left(\frac{\Phi + 2k\pi}{n}\right)} $$The \(n\) distinct roots are found by substituting \(k = 0, 1, 2, \dots, n-1\).
4.3. Step-by-Step Process for Finding Roots
Let's find the cube roots of \(w = 8i\).
- Express \(w\) in Exponential Form:
\(|w| = R = 8\).
\(8i\) lies on the positive imaginary axis, so the principal argument is \(\Phi = \pi/2\).
\(w = 8\text{e}^{i\pi/2}\). - Write the General Form of \(w\):
\(w = 8\text{e}^{i(\pi/2 + 2k\pi)}\). (This is the essential step!) - Apply the Root Formula (\(n=3\)):
\(z_k = \sqrt[3]{8} \text{e}^{i\left(\frac{\pi/2 + 2k\pi}{3}\right)}\)
\(z_k = 2 \text{e}^{i\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)}\) - Calculate Roots for \(k=0, 1, 2\):
k = 0: \(z_0 = 2 \text{e}^{i(\pi/6)}\)
k = 1: \(z_1 = 2 \text{e}^{i(\pi/6 + 2\pi/3)} = 2 \text{e}^{i(5\pi/6)}\)
k = 2: \(z_2 = 2 \text{e}^{i(\pi/6 + 4\pi/3)} = 2 \text{e}^{i(9\pi/6)} = 2 \text{e}^{i(3\pi/2)}\) - Convert back (optional, but often required):
\(z_0 = 2(\cos(\pi/6) + i\sin(\pi/6)) = \sqrt{3} + i\)
\(z_2 = 2(\cos(3\pi/2) + i\sin(3\pi/2)) = -2i\)
4.4. Geometry of the Roots
When plotted on an Argand diagram, the \(n\) roots of a complex number always share the following properties:
- They all lie on a circle centered at the origin with radius \(\sqrt[n]{R}\).
- They are equally spaced around this circle. The angle between consecutive roots is exactly \(2\pi/n\) (or \(360^\circ/n\)).
Analogy: Imagine the roots are equally spaced seats on a Ferris wheel, all the same distance from the center. For the cube roots above, they are separated by \(2\pi/3 = 120^\circ\).
4.5. Roots of Unity
A special case is finding the roots of unity, which means solving \(z^n = 1\).
Since \(1 = 1\text{e}^{i0}\), the roots are:
$$ z_k = 1 \text{e}^{i\left(\frac{2k\pi}{n}\right)} $$If we denote the first non-real root (when \(k=1\)) as \(\omega = \text{e}^{i(2\pi/n)}\), then all the roots are: \(1, \omega, \omega^2, \omega^3, \dots, \omega^{n-1}\).
Common Mistake to Avoid: DO NOT forget to add the \(2k\pi\) to the argument \(\Phi\) before dividing by \(n\). If you forget this step, you will only find one root, not the required \(n\) roots!
Key Takeaway: The periodicity of the argument (\(\theta \equiv \theta + 2k\pi\)) is necessary to find all \(n\) distinct roots, which are always equally spaced on the Argand diagram.
5. Connecting Complex Numbers to Trigonometry
De Moivre's Theorem is a bridge between powers of complex numbers and multiple angle identities (like \(\cos(3\theta)\) or \(\sin(4\theta)\)).
5.1. Deriving Identities
To express \(\cos(n\theta)\) or \(\sin(n\theta)\) in terms of powers of \(\cos\theta\) and \(\sin\theta\):
- Start with De Moivre: \(\cos(n\theta) + i\sin(n\theta) = (\cos\theta + i\sin\theta)^n\).
- Expand the right-hand side using the Binomial Theorem.
- Group the terms containing \(i\) (Imaginary part) and the terms without \(i\) (Real part).
- Equate the corresponding parts:
- Real part = \(\cos(n\theta)\)
- Imaginary part = \(\sin(n\theta)\)
Example: Finding \(\cos(3\theta)\)
\(\cos(3\theta) + i\sin(3\theta) = (\cos\theta + i\sin\theta)^3\)
\(= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3\)
Recall \(i^2 = -1\) and \(i^3 = -i\):
\(= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta\)
Equating Real Parts:
\(\cos(3\theta) = \cos^3\theta - 3\cos\theta\sin^2\theta\). (We have successfully derived the triple angle formula!)
Accessibility Tip: The algebra here can be intense. Use a new line for every term in the binomial expansion and substitute the powers of \(i\) immediately (\(i^2 \to -1\), \(i^3 \to -i\)). Use brackets heavily to keep real and imaginary parts separated until the very end.
Key Takeaway: De Moivre’s theorem combined with the binomial expansion allows us to derive trigonometric identities for multiple angles by simply equating the real or imaginary components.
🎉 Chapter Summary: Your FP2 Complex Numbers Toolkit
You now have the tools needed to handle complex numbers at a higher level:
- De Moivre's Theorem: \((r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))\). Use this for integer powers and for deriving trigonometric identities.
- Exponential Form: \(z = r\text{e}^{i\theta}\). Use this for efficient multiplication, division, and especially when finding roots.
- Finding \(n\)-th Roots: Always start by writing the complex number \(w\) in its general argument form: \(w = R\text{e}^{i(\Phi + 2k\pi)}\). This ensures you find all \(n\) equally spaced solutions.
You've tackled some of the most abstract parts of Further Pure Mathematics. Keep practicing those root questions—they are the most technical skill in this chapter. Good luck!