Complex numbers can sound mysterious, but they were invented to solve a very practical problem: equations like \(x^2 + 1 = 0\) that have no real solutions. Today they are essential in FSc and A Level maths, engineering, physics and electronics. This guide explains what complex numbers are, how to calculate with them, and how to picture them, with step-by-step examples.
The Imaginary Unit \(i\)
No real number squares to give a negative result. So mathematicians defined a new number, \(i\), with
$$ i^2 = -1 \qquad \text{or} \qquad i = \sqrt{-1} $$
This lets us take square roots of negative numbers:
$$ \sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i $$
Powers of \(i\)
The powers of \(i\) repeat in a cycle of four:
$$ i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1 $$
To find \(i^n\), divide \(n\) by 4 and use the remainder. For example, \(i^{27}\): \(27 = 4 \times 6 + 3\), so \(i^{27} = i^3 = -i\).
What Is a Complex Number?
A complex number has the form
$$ z = a + bi $$
where \(a\) and \(b\) are real numbers.
- \(a\) is the real part, written \(\text{Re}(z)\).
- \(b\) is the imaginary part, written \(\text{Im}(z)\).
For \(z = 3 - 5i\), \(\text{Re}(z) = 3\) and \(\text{Im}(z) = -5\). Note that the imaginary part is the real number \(-5\), not \(-5i\).
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal.
Arithmetic with Complex Numbers
Addition and Subtraction
Combine real parts and imaginary parts separately:
$$ (4 + 3i) + (2 - 7i) = 6 - 4i $$
Multiplication
Expand like brackets, then replace \(i^2\) with \(-1\).
Worked Example 1
Calculate \((2 + 3i)(4 - i)\).
Step 1: Expand:
$$ 2(4) + 2(-i) + 3i(4) + 3i(-i) = 8 - 2i + 12i - 3i^2 $$
Step 2: Replace \(i^2\) with \(-1\):
$$ 8 + 10i - 3(-1) = 8 + 10i + 3 $$
Step 3: Simplify:
$$ (2 + 3i)(4 - i) = 11 + 10i $$
The Complex Conjugate
The conjugate of \(z = a + bi\) is \(\bar{z} = a - bi\). Multiplying a number by its conjugate always gives a real number:
$$ (a + bi)(a - bi) = a^2 + b^2 $$
Division
To divide, multiply the top and bottom by the conjugate of the denominator.
Worked Example 2
Simplify \(\dfrac{5 + i}{2 - 3i}\).
Step 1: The conjugate of \(2 - 3i\) is \(2 + 3i\). Multiply top and bottom:
$$ \frac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} $$
Step 2: Numerator:
$$ 10 + 15i + 2i + 3i^2 = 10 + 17i - 3 = 7 + 17i $$
Step 3: Denominator:
$$ 2^2 + 3^2 = 4 + 9 = 13 $$
Step 4: Write in \(a + bi\) form:
$$ \frac{5 + i}{2 - 3i} = \frac{7}{13} + \frac{17}{13}i $$
Check: \(\left(\frac{7}{13} + \frac{17}{13}i\right)(2 - 3i) = \frac{14 - 21i + 34i - 51i^2}{13} = \frac{14 + 51 + 13i}{13} = \frac{65 + 13i}{13} = 5 + i\). Correct.
The Argand Diagram
A complex number \(a + bi\) can be plotted as the point \((a, b)\) on a plane, with the real part on the horizontal axis and the imaginary part on the vertical axis. This is called the Argand diagram.
Modulus
The modulus is the distance from the origin:
$$ |z| = \sqrt{a^2 + b^2} $$
Argument
The argument, \(\arg z = \theta\), is the angle the line makes with the positive real axis, usually taken in the range \(-\pi < \theta \leq \pi\). Use \(\tan \theta = \frac{b}{a}\), but always check the quadrant from a sketch.
Polar (Modulus-Argument) Form
Any complex number can be written as
$$ z = r(\cos\theta + i\sin\theta) $$
where \(r = |z|\) and \(\theta = \arg z\).
Worked Example 3
Write \(z = -1 + \sqrt{3}\,i\) in polar form.
Step 1: Find the modulus:
$$ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 $$
Step 2: Find the argument. The point \((-1, \sqrt{3})\) is in the second quadrant. The reference angle is
$$ \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} $$
so \(\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).
Step 3: Write the polar form:
$$ z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) $$
Check: \(2\cos\frac{2\pi}{3} = 2\left(-\frac{1}{2}\right) = -1\) and \(2\sin\frac{2\pi}{3} = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}\). Correct.
Solving Quadratics with Complex Roots
Worked Example 4
Solve \(x^2 - 4x + 13 = 0\).
Step 1: Discriminant: \(D = (-4)^2 - 4(1)(13) = 16 - 52 = -36\).
Step 2: Apply the quadratic formula:
$$ x = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i $$
The roots \(2 + 3i\) and \(2 - 3i\) are conjugates, which always happens for real-coefficient quadratics.
Common Mistakes
- Forgetting \(i^2 = -1\) after expanding brackets.
- Writing the imaginary part as \(bi\) instead of \(b\).
- Using \(\tan^{-1}\left(\frac{b}{a}\right)\) blindly for the argument without checking the quadrant.
- Leaving \(i\) in the denominator. Always multiply by the conjugate.
Key Takeaways
- \(i = \sqrt{-1}\), and its powers repeat every four: \(i, -1, -i, 1\).
- A complex number \(a + bi\) has a real part \(a\) and an imaginary part \(b\).
- Divide by multiplying top and bottom by the conjugate of the denominator.
- On the Argand diagram, \(|z| = \sqrt{a^2 + b^2}\) and the argument gives the angle, leading to the polar form \(r(\cos\theta + i\sin\theta)\).
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