Welcome to Complex Numbers! (Unit FP1)
Hello! You've successfully navigated the foundations of Pure Mathematics, and now we’re diving into a genuinely fascinating topic: Complex Numbers. Don't worry if this feels a little strange at first—you are moving from the world of numbers you can easily visualize (the real number line) into a new, two-dimensional mathematical space.
In this chapter, we will learn how to handle the impossible (finding the square root of a negative number!), how to perform basic algebra with these new quantities, and how to visualize them using geometry. Complex numbers are vital tools used everywhere, from understanding alternating current (AC) electricity to designing suspension bridges. Let's get started!
1. Defining the Imaginary Unit and Complex Numbers
1.1 The Birth of \(i\)
For centuries, mathematicians knew that equations like \(x^2 = -4\) had no solutions within the Real Number System (\(\mathbb{R}\)). To fix this, we created a new number, the imaginary unit, \(i\).
- The defining property of the imaginary unit is:
$${i^2 = -1}$$ - This means: $${i = \sqrt{-1}}$$
Analogy: Think of \(i\) as a special type of currency. It doesn't exist in your bank account (Real numbers), but it’s essential for mathematical trade!
1.2 What is a Complex Number?
A Complex Number, usually denoted by the variable \(z\), is any number that can be written in the form:
$${z = a + bi}$$where \(a\) and \(b\) are real numbers.
- \(a\) is the Real Part, denoted by \(\text{Re}(z)\).
- \(b\) is the Imaginary Part, denoted by \(\text{Im}(z)\).
Quick Check: If \(z = 3 - 5i\), then \(\text{Re}(z) = 3\) and \(\text{Im}(z) = -5\). (Note: The imaginary part is just the coefficient \(b\), not \(bi\) itself!)
Powers of \(i\): The 4-Cycle Rule
The powers of \(i\) repeat in a cycle of four. This is a brilliant memory aid for simplifying complex calculations.
- $${i^1 = i}$$
- $${i^2 = -1}$$
- $${i^3 = i^2 \times i = -1 \times i = -i}$$
- $${i^4 = i^2 \times i^2 = (-1) \times (-1) = 1}$$
The cycle starts again at \(i^5 = i\). To calculate any high power of \(i\), divide the exponent by 4 and look at the remainder.
Quick Review: The Foundation
- $${i^2 = -1}$$
- $${z = a + bi}$$ (Standard form)
- If \(i\) has an exponent divisible by 4 (remainder 0), the result is 1.
2. Arithmetic of Complex Numbers
Performing calculations with complex numbers in the form \(a + bi\) is very similar to working with algebraic expressions involving a variable \(x\), with one crucial difference: remember that $${i^2 = -1}$$.
2.1 Addition and Subtraction
Treat the Real parts and the Imaginary parts separately, just like collecting like terms.
If \(z_1 = a + bi\) and \(z_2 = c + di\):
$${z_1 + z_2 = (a+c) + (b+d)i}$$
Example: If \(z_1 = 2 + 3i\) and \(z_2 = 5 - 7i\).
$$z_1 + z_2 = (2+5) + (3-7)i = 7 - 4i$$
2.2 Multiplication
Use the distributive law (or the FOIL method, like multiplying two binomials).
Step-by-step Example: Multiply \(z_1 = 3 + 2i\) by \(z_2 = 1 - 4i\).
- Expand using FOIL: $$(3 + 2i)(1 - 4i) = 3(1) - 3(4i) + 2i(1) - 2i(4i)$$ $$= 3 - 12i + 2i - 8i^2$$
- Group the \(i\) terms: $$= 3 - 10i - 8i^2$$
- Substitute \(i^2 = -1\): $$= 3 - 10i - 8(-1)$$ $$= 3 - 10i + 8$$
- Combine the Real parts: $$= 11 - 10i$$
Key Takeaway for Multiplication: Always look out for the \(i^2\) terms and immediately turn them into Real numbers!
3. The Complex Conjugate and Division
Division is the trickiest of the basic operations, but we have a powerful tool to make it simple: the Complex Conjugate.
3.1 Defining the Conjugate
The conjugate of a complex number \(z = a + bi\) is written as \(\bar{z}\) (sometimes \(z^*\)). You find the conjugate simply by changing the sign of the imaginary part.
If \(z = a + bi\), then \(\bar{z} = a - bi\).
Example: If \(z = 5 - 3i\), then \(\bar{z} = 5 + 3i\).
Why is the Conjugate Useful?
When you multiply a complex number by its conjugate, the imaginary terms cancel out, leaving a purely real result:
$$(a + bi)(a - bi) = a^2 - abi + abi - b^2i^2$$ $$= a^2 - b^2(-1)$$ $$= a^2 + b^2$$
This is always a non-negative real number!
3.2 Division
When dividing complex numbers, our goal is to eliminate the \(i\) from the denominator (a process similar to rationalising a surd). We do this by multiplying the numerator AND the denominator by the conjugate of the denominator.
Step-by-step Example: Find \(\frac{2+i}{3-4i}\).
- Identify the denominator and its conjugate.
Denominator: \(3 - 4i\). Conjugate: \(3 + 4i\). - Multiply the fraction by \(\frac{3+4i}{3+4i}\):
$$\frac{2+i}{3-4i} \times \frac{3+4i}{3+4i}$$ - Calculate the denominator (this is the easy part, \(a^2 + b^2\)):
$$(3-4i)(3+4i) = 3^2 + 4^2 = 9 + 16 = 25$$ - Calculate the numerator (using FOIL):
$$(2+i)(3+4i) = 6 + 8i + 3i + 4i^2$$ $$= 6 + 11i - 4 = 2 + 11i$$ - Combine the results and write in standard \(a + bi\) form:
$$\frac{2 + 11i}{25} = \frac{2}{25} + \frac{11}{25}i$$
Common Mistake to Avoid: When dividing, ensure you use the conjugate of the denominator, not the numerator!
4. Solving Polynomial Equations
One of the main reasons complex numbers exist is to ensure that polynomial equations always have solutions. In FP1, we focus on polynomials (quadratics, cubics, and quartics) with real coefficients.
4.1 The Conjugate Root Theorem (Crucial FP1 Concept)
If a polynomial equation, \(P(x) = 0\), has all real coefficients, and if \(z = a + bi\) is a root, then its conjugate, \(\bar{z} = a - bi\), must also be a root.
Why is this important? It means complex roots always come in pairs. If you find one complex root, you immediately know a second one for free! This is critical for solving cubic and quartic equations.
4.2 Solving a Cubic Equation
A cubic equation (\(ax^3 + bx^2 + cx + d = 0\)) must have 3 roots. Since coefficients are real, there are only two possibilities:
- Three real roots.
- One real root and a pair of complex conjugate roots.
Step-by-step Example: A cubic equation \(P(x) = 0\) has real coefficients. Given that \(2\) and \(3 + i\) are two roots, find the third root.
- Identify the complex root: \(z_1 = 3 + i\).
- Since the coefficients are real, the conjugate root theorem applies.
- The second complex root must be the conjugate: \(z_2 = 3 - i\).
- The three roots are \(2, 3 + i,\) and \(3 - i\). (The real root 2 is its own conjugate.)
4.3 Finding the Polynomial from its Roots
If you know the roots, you can find the polynomial by multiplying the factors \((x - \text{root})\).
If \(z_1\) and \(\bar{z_1}\) are a pair of conjugate roots, it is smart to multiply their factors first:
$$(x - z_1)(x - \bar{z_1})$$
Let \(z_1 = a + bi\).
$$(x - (a + bi))(x - (a - bi))$$
When you expand this, the imaginary terms cancel out, always resulting in a real quadratic factor:
$$(x^2 - (z_1 + \bar{z_1})x + z_1 \bar{z_1})$$
Remember:
- Sum of roots: \(z_1 + \bar{z_1} = (a+bi) + (a-bi) = 2a\) (Twice the real part)
- Product of roots: \(z_1 \bar{z_1} = a^2 + b^2\) (Modulus squared - see Section 5)
Therefore, the quadratic factor for the pair \(a \pm bi\) is always:
$${x^2 - 2ax + (a^2 + b^2)}$$Memory Aid: If you have roots \(1 \pm 2i\), then \(a=1, b=2\). The factor is \(x^2 - 2(1)x + (1^2 + 2^2) = x^2 - 2x + 5\). Much faster than FOILing four separate terms!
Did You Know?
The Fundamental Theorem of Algebra states that a polynomial of degree \(n\) (where \(n \ge 1\)) with complex coefficients must have exactly \(n\) roots (counted with multiplicity) in the complex number system. This guarantees all polynomials can be solved!
5. Visualising Complex Numbers: The Argand Diagram
Since a complex number \(z = a + bi\) has two components (\(a\) and \(b\)), we cannot plot it on a simple number line. Instead, we use a two-dimensional plane called the Argand Diagram.
5.1 The Complex Plane
The Argand diagram is essentially a Cartesian coordinate system, where:
- The horizontal axis is the Real Axis (representing \(a\)).
- The vertical axis is the Imaginary Axis (representing \(b\)).
A complex number \(z = a + bi\) is plotted as the coordinate point \((a, b)\).
Example: \(z = 4 - 3i\) is plotted at the point \((4, -3)\) in the fourth quadrant.
5.2 The Modulus: Distance from the Origin
The Modulus of \(z\), written as \(|z|\), represents the distance of the point \((a, b)\) from the origin \((0, 0)\). Since the axes are perpendicular, we use Pythagoras' Theorem.
$${|z| = \sqrt{a^2 + b^2}}$$Since \(a\) and \(b\) are squared, the modulus is always a non-negative real number.
Example: Find the modulus of \(z = -5 + 12i\).
$$|z| = \sqrt{(-5)^2 + (12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$
5.3 The Argument: The Angle
The Argument of \(z\), written as \(\arg(z)\) or \(\theta\), is the angle between the positive real axis and the line segment connecting the origin to the point \(z\).
In FP1, we generally work with the Principal Argument, which must satisfy:
$${-\pi < \theta \le \pi \quad \text{ or } \quad -180^\circ < \theta \le 180^\circ}$$(Depending on whether radians or degrees are required.)
Finding the Argument (Step-by-Step)
- Sketch It! This is the most important step. Draw the Argand diagram to identify which quadrant \(z\) lies in.
- Find the Reference Angle ($\alpha$). Use trigonometry based on the lengths \(|a|\) and \(|b|\) (the non-negative lengths of the sides of the triangle). $${\tan(\alpha) = \frac{|b|}{|a|}}$$
- Find the Principal Argument ($\theta$). Adjust \(\alpha\) based on the quadrant:
- Quadrant I (\(a>0, b>0\)): \(\theta = \alpha\)
- Quadrant II (\(a<0, b>0\)): \(\theta = \pi - \alpha\)
- Quadrant III (\(a<0, b<0\)): \(\theta = -(\pi - \alpha)\) (Moving backwards from the positive real axis)
- Quadrant IV (\(a>0, b<0\)): \(\theta = -\alpha\)
Example (Radians): Find the argument of \(z = -1 - i\).
- Sketch: \(z\) is in the third quadrant. \(a=-1, b=-1\).
- Reference Angle ($\alpha$): \(\tan(\alpha) = \frac{|-1|}{|-1|} = 1\). Therefore, \(\alpha = \frac{\pi}{4}\).
- Principal Argument ($\theta$): Since it's in Quadrant III, we calculate \(\pi - \alpha\), then make it negative. $$\theta = - \left(\pi - \frac{\pi}{4}\right) = - \frac{3\pi}{4}$$
Key Takeaway for Arguments: Never calculate the argument directly using \(\arctan(b/a)\) without checking the quadrant first. The calculator only gives you angles in Q1 or Q4!
Conclusion: You've Mastered the Basics!
You now have a solid foundation in the complex number system for FP1! You can handle the algebra, solve challenging polynomials using the conjugate root theorem, and understand the geometric representation via the Argand diagram, modulus, and argument. Keep practicing the division and argument calculation steps—they are where most students make small errors. Good luck!