👋 Welcome to the World of Complex Numbers!

Don't worry if the name "Complex Numbers" sounds intimidating—it just means they involve the Imaginary Unit. In fact, these numbers are essential because they allow us to solve any polynomial equation, something real numbers alone cannot do!

This chapter is a foundational pillar of Further Pure Mathematics. By the end of these notes, you'll be able to confidently handle these numbers both algebraically and geometrically. Let's dive in!

Section 1: Introducing the Imaginary Unit and Basic Operations

1.1 Defining the Imaginary Unit (\(i\))

For centuries, mathematicians couldn't solve equations like \(x^2 + 1 = 0\). If you try to rearrange it, you get \(x^2 = -1\). In the realm of real numbers, this is impossible because squaring any real number (positive or negative) results in a positive number.

To fix this, we introduce the Imaginary Unit, denoted by \(i\):

  • Definition: \(i\) is the number such that \(i^2 = -1\).
  • Therefore, \(i = \sqrt{-1}\).

1.2 What is a Complex Number?

A complex number, usually denoted by \(z\), is a combination of a real part and an imaginary part.

$$z = a + bi$$

  • \(a\) is the Real Part (\(\text{Re}(z)\)).
  • \(b\) is the Imaginary Part (\(\text{Im}(z)\)). Note: The imaginary part is just the coefficient \(b\), not \(bi\).

Example: If \(z = 3 + 4i\), then \(\text{Re}(z) = 3\) and \(\text{Im}(z) = 4\).

1.3 Adding and Subtracting Complex Numbers

This is the easiest part! You treat \(i\) like a variable (like \(x\)), combining only the real parts together and the imaginary parts together.

If \(z_1 = a + bi\) and \(z_2 = c + di\):

  • Addition: \(z_1 + z_2 = (a+c) + (b+d)i\)
  • Subtraction: \(z_1 - z_2 = (a-c) + (b-d)i\)

1.4 Multiplying Complex Numbers

Multiplication works exactly like multiplying two brackets in algebra (using the FOIL method), but you must substitute \(i^2 = -1\) whenever it appears.

If \(z_1 = (2 + 3i)\) and \(z_2 = (4 - i)\):

$$z_1 z_2 = (2 + 3i)(4 - i)$$

  1. Multiply (FOIL): \(8 - 2i + 12i - 3i^2\)
  2. Substitute \(i^2 = -1\): \(8 + 10i - 3(-1)\)
  3. Simplify: \(8 + 10i + 3 = 11 + 10i\)

🔥 Key Takeaway: Complex numbers are defined by \(i^2 = -1\). Operations are simple: add/subtract like terms; multiply using FOIL and replace \(i^2\) with \(-1\).

Section 2: The Complex Conjugate and Division

2.1 The Complex Conjugate (\(z^*\))

The complex conjugate of a number \(z\) is found by simply changing the sign of the imaginary part. It is denoted by \(z^*\) or \(\bar{z}\).

  • If \(z = a + bi\), then \(z^* = a - bi\).

Example: If \(z = 5 - 2i\), then \(z^* = 5 + 2i\). If \(w = -3i\), then \(w^* = 3i\).

Why is the Conjugate Important?

Multiplying a complex number by its conjugate always results in a purely real number.

$$z z^* = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 - b^2(-1) = a^2 + b^2$$

This is extremely useful for division!

2.2 Dividing Complex Numbers

You cannot leave \(i\) in the denominator of a fraction. Division is achieved by a process similar to rationalizing the denominator with surds: multiplying the numerator and the denominator by the conjugate of the denominator.

Step-by-Step Division Example

Calculate \(\frac{1 + 2i}{3 - 4i}\).

  1. Identify the conjugate: The denominator is \(3 - 4i\). Its conjugate is \(3 + 4i\).
  2. Multiply top and bottom: $$\frac{1 + 2i}{3 - 4i} \times \frac{3 + 4i}{3 + 4i}$$
  3. Calculate the Denominator: (The easy part: use \(a^2 + b^2\)) $$3^2 + 4^2 = 9 + 16 = 25$$
  4. Calculate the Numerator: (Use FOIL) $$(1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i^2$$ $$3 + 10i + 8(-1) = 3 + 10i - 8 = -5 + 10i$$
  5. Combine and Simplify: $$\frac{-5 + 10i}{25} = \frac{-5}{25} + \frac{10}{25}i = -\frac{1}{5} + \frac{2}{5}i$$

⚠️ Common Mistake: Students often forget to multiply the numerator by the conjugate as well! Always multiply by \(\frac{z^*}{z^*}\) (which is essentially 1).

🔥 Key Takeaway: Use the conjugate \(z^*\) to perform division. It ensures the denominator becomes a real number (\(a^2 + b^2\)).

Section 3: Solving Polynomial Equations

3.1 Complex Roots of Quadratic Equations

When solving a quadratic equation \(ax^2 + bx + c = 0\) (where \(a, b, c\) are real numbers) using the quadratic formula, if the discriminant (\(b^2 - 4ac\)) is negative, you will generate complex roots.

Example: Solve \(z^2 + 2z + 5 = 0\).

Using the formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

$$z = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2(1)}$$ $$z = \frac{-2 \pm \sqrt{4 - 20}}{2}$$ $$z = \frac{-2 \pm \sqrt{-16}}{2}$$

Since \(\sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i\):

$$z = \frac{-2 \pm 4i}{2}$$ $$z = -1 \pm 2i$$

The roots are \(z_1 = -1 + 2i\) and \(z_2 = -1 - 2i\). Notice something important...

3.2 The Conjugate Root Theorem

For any polynomial equation with real coefficients, if one root is a complex number, \(z\), then its conjugate, \(z^*\), must also be a root.

This is incredibly helpful for higher-order polynomials (cubics and quartics).

  • If \(3 + i\) is a root of a cubic equation, then \(3 - i\) must also be a root.

3.3 Finding Unknown Roots and Coefficients

If you are given one complex root of a cubic equation \(P(z) = 0\), you can use the Conjugate Root Theorem to find the quadratic factor associated with that pair of roots.

If the roots are \(z_1\) and \(z_1^*\), then the quadratic factor is: $$(z - z_1)(z - z_1^*) = z^2 - (\text{sum of roots})z + (\text{product of roots})$$

Since \(z_1 + z_1^*\) and \(z_1 z_1^*\) are both real numbers, this quadratic factor will have real coefficients.

Step-by-Step for Cubics:

  1. Identify the given root \(z_1\) and its conjugate \(z_1^*\).
  2. Calculate the Sum (\(S\)) and Product (\(P\)) of these two roots.
  3. Form the quadratic factor: \(Q(z) = z^2 - Sz + P\).
  4. Divide the original cubic polynomial \(P(z)\) by \(Q(z)\) to find the remaining linear factor \((z - k)\).
  5. The real root is \(z = k\).

🔥 Key Takeaway: Complex roots of polynomials with real coefficients always come in conjugate pairs. Use this theorem to find the factors and solve the equation.

Section 4: The Argand Diagram (Geometry)

4.1 Plotting Complex Numbers

The Argand Diagram (or Complex Plane) is the graphical representation of complex numbers. Instead of an x-y coordinate system, we use axes representing the Real and Imaginary parts.

  • The horizontal axis is the Real Axis (\(x\)).
  • The vertical axis is the Imaginary Axis (\(y\)).

A complex number \(z = a + bi\) is plotted as the point \((a, b)\).

Example: Plot \(z_1 = 4 + i\) and \(z_2 = -2 + 3i\).
\(z_1\) is plotted at \((4, 1)\). \(z_2\) is plotted at \((-2, 3)\).

4.2 Visualizing Operations

When you add or subtract complex numbers on the Argand Diagram, it behaves exactly like vector addition or subtraction.

  • If you plot the vector from the origin to \(z_1\) and the vector from the origin to \(z_2\), the sum \(z_1 + z_2\) is found using the parallelogram rule.
  • The conjugate \(z^*\) is the reflection of \(z\) across the Real Axis.

Did you know? This geometric interpretation is why complex numbers are so powerful in physics and engineering—they represent both magnitude and direction simultaneously!

Section 5: Modulus and Argument (Polar Coordinates)

To describe a complex number geometrically, we can use its Cartesian coordinates \((a, b)\), or we can use Polar Coordinates: its distance from the origin (\(r\)) and the angle it makes with the positive real axis (\(\theta\)).

5.1 The Modulus (\(|z|\))

The Modulus, denoted \(|z|\) or \(r\), is the distance from the origin to the point \(z\) on the Argand Diagram. Using Pythagoras’ theorem:

$$|z| = r = \sqrt{a^2 + b^2}$$

Note that \(|z|^2 = z z^* = a^2 + b^2\).

Example: If \(z = 3 - 4i\), then \(|z| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).

5.2 The Argument (\(\arg(z)\))

The Argument, denoted \(\arg(z)\) or \(\theta\), is the angle between the positive Real Axis and the line segment connecting the origin to \(z\).

Convention (FP1 Standard): The argument is usually measured in radians and must lie in the range:

$$-\pi < \theta \leq \pi$$

Calculating the Argument (Step-by-Step)

This is the trickiest part for many students. Do NOT just use \(\arctan(b/a)\). You must use the Argand diagram to find the correct quadrant first.

Step 1: Find the Reference Angle (\(\alpha\)).
Use the positive lengths of the sides (ignore the signs of \(a\) and \(b\)): $$\alpha = \arctan\left(\left|\frac{b}{a}\right|\right)$$

Step 2: Determine the Quadrant and Adjust \(\theta\).

  • Q1 (\(a>0, b>0\)): \(\theta = \alpha\)
  • Q2 (\(a<0, b>0\)): \(\theta = \pi - \alpha\)
  • Q3 (\(a<0, b<0\)): \(\theta = -\pi + \alpha\) (or \(\alpha - \pi\))
  • Q4 (\(a>0, b<0\)): \(\theta = -\alpha\)

Analogy: Think of \(\alpha\) as your calculation result, and the quadrant adjustment makes sure the angle is measured correctly from the positive x-axis, staying within the \(-\pi\) to \(\pi\) range.

Example: Find the argument of \(z = -1 - i\).

  1. \(a=-1\), \(b=-1\). This is in Quadrant 3.
  2. Reference Angle: \(\alpha = \arctan\left(\left|\frac{-1}{-1}\right|\right) = \arctan(1) = \frac{\pi}{4}\).
  3. Adjust for Q3: \(\theta = -\pi + \alpha = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}\).

🔥 Key Takeaway: Modulus (\(r\)) is distance (\(\sqrt{a^2+b^2}\)). Argument (\(\theta\)) is the angle, calculated using arctan and crucially adjusting for the correct quadrant (Q1, Q2, Q3, Q4) to keep \(\theta \in (-\pi, \pi]\).

Section 6: Modulus-Argument Form (Polar Form)

6.1 The Definition

Once you have the modulus \(r\) and the argument \(\theta\), you can express the complex number \(z = a + bi\) in its Modulus-Argument Form (also called the Polar Form).

Look at the diagram:

  • The real part \(a\) is adjacent to the angle \(\theta\): \(a = r \cos \theta\)
  • The imaginary part \(b\) is opposite the angle \(\theta\): \(b = r \sin \theta\)

Substituting these back into \(z = a + bi\) gives:

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

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

This form is extremely powerful for later calculations involving powers and roots (though those specific techniques are typically covered in FP2).

6.2 Converting Between Forms

A) Cartesian to Polar (\(a+bi\) to \(r(\cos \theta + i \sin \theta)\))

(This is simply Section 5 combined.)

  1. Calculate \(r = \sqrt{a^2 + b^2}\).
  2. Calculate \(\theta\) (argument), ensuring it's in the correct range \(-\pi < \theta \leq \pi\).

Example: Convert \(z = \sqrt{3} + i\) to polar form.

  1. \(r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2\).
  2. \(a>0, b>0\). Q1. \(\alpha = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\).
  3. Polar Form: \(z = 2\left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)\).
B) Polar to Cartesian (\(r(\cos \theta + i \sin \theta)\) to \(a+bi\))

This is much easier! Just evaluate the cosine and sine values and expand the bracket.

Example: Convert \(z = 4\left(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}\right)\) to Cartesian form.

We know \(\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}\) and \(\sin \frac{3\pi}{4} = \frac{1}{\sqrt{2}}\).

$$z = 4\left(-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}\right)$$ $$z = -\frac{4}{\sqrt{2}} + i \frac{4}{\sqrt{2}}$$ $$z = -2\sqrt{2} + i 2\sqrt{2}$$

🎉 Success Check! If you can move fluently between \(a+bi\) and \(r(\cos \theta + i \sin \theta)\) and handle all four basic operations, you have mastered the essentials of FP1 Complex Numbers!