Complex Numbers – Part 1
Introduction, definitions and basic operations
🌟 Why Do We Need Complex Numbers?
Let's start with a simple question:
What is the solution of the equation \(x^2 = -1\) ?
In the real numbers there is no solution – because the square of any real number is non-negative!
To solve problems like this, a new number called i (from Imaginary) was invented.
⭐ Definition of the Imaginary Unit i
\(i = \sqrt{-1}\)
\(i^2 = -1\)
💡 Now there is a solution!
\(x^2 = -1\)
\(x = \pm\sqrt{-1} = \pm i\)
🔄 Powers of i (cyclic!)
\(i^0 = 1\)
\(i^1 = i\)
\(i^2 = -1\)
\(i^3 = -i\)
Then it repeats from the beginning!
\(i^4 = 1, \quad i^5 = i, \quad i^6 = -1, \quad i^7 = -i, \quad ...\)
🔑 How to compute \(i^n\)?
Divide n by 4 and look at the remainder:
- Remainder 0 → \(i^n = 1\)
- Remainder 1 → \(i^n = i\)
- Remainder 2 → \(i^n = -1\)
- Remainder 3 → \(i^n = -i\)
Example: \(i^{23} = ?\)
\(23 \div 4 = 5\) remainder \(3\)
Therefore: \(i^{23} = i^3 = -i\)
📐 What Is a Complex Number?
\(z = a + bi\)
| \(z\) | The complex number |
| \(a\) | Real part – denoted \(\text{Re}(z)\) |
| \(b\) | Imaginary part – denoted \(\text{Im}(z)\) |
| \(i\) | The imaginary unit (\(i^2 = -1\)) |
Examples:
| \(z = 3 + 2i\) | → \(a = 3, \, b = 2\) |
| \(z = -1 + 4i\) | → \(a = -1, \, b = 4\) |
| \(z = 5 - 3i\) | → \(a = 5, \, b = -3\) |
| \(z = 7\) | → \(a = 7, \, b = 0\) (a real number!) |
| \(z = 4i\) | → \(a = 0, \, b = 4\) (purely imaginary) |
⚖️ Equality of Complex Numbers
Two complex numbers are equal if and only if:
\(a + bi = c + di \iff a = c \text{ and } b = d\)
💡 In words: the real parts are equal and the imaginary parts are equal.
Example: find x and y if \(2x + 3yi = 6 - 9i\)
Equating real parts: \(2x = 6 \implies x = 3\)
Equating imaginary parts: \(3y = -9 \implies y = -3\)
➕ Addition and Subtraction
\((a + bi) + (c + di) = (a + c) + (b + d)i\)
\((a + bi) - (c + di) = (a - c) + (b - d)i\)
💡 The rule: add/subtract real parts separately and imaginary parts separately.
Examples:
\((3 + 2i) + (1 + 4i) = (3+1) + (2+4)i = 4 + 6i\)
\((5 - 3i) - (2 + i) = (5-2) + (-3-1)i = 3 - 4i\)
\((7 + 2i) + (-7 + 3i) = 0 + 5i = 5i\)
✖️ Multiplication
\((a + bi)(c + di) = (ac - bd) + (ad + bc)i\)
💡 How to remember? Expand brackets normally and use \(i^2 = -1\):
\((a + bi)(c + di)\)
\(= ac + adi + bci + bdi^2\)
\(= ac + adi + bci + bd(-1)\)
\(= (ac - bd) + (ad + bc)i\)
Example: \((2 + 3i)(4 - i)\)
\(= 2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i)\)
\(= 8 - 2i + 12i - 3i^2\)
\(= 8 - 2i + 12i - 3(-1)\)
\(= 8 + 3 + (-2 + 12)i\)
\(= 11 + 10i\)
🪞 Conjugate
If \(z = a + bi\), then its conjugate is:
\(\bar{z} = a - bi\)
💡 In words: flip the sign of the imaginary part only!
Examples:
| \(z = 3 + 2i\) | → | \(\bar{z} = 3 - 2i\) |
| \(z = 5 - 4i\) | → | \(\bar{z} = 5 + 4i\) |
| \(z = -2i\) | → | \(\bar{z} = 2i\) |
| \(z = 7\) | → | \(\bar{z} = 7\) (a real number equals its own conjugate) |
⭐ Important property:
\(z \cdot \bar{z} = a^2 + b^2\)
(always a non-negative real number!)
Proof:
\((a + bi)(a - bi) = a^2 - abi + abi - b^2i^2 = a^2 - b^2(-1) = a^2 + b^2\)
➗ Division
To divide, multiply numerator and denominator by the conjugate of the denominator:
\(\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}\)
Example: \(\frac{3 + 2i}{1 - i}\)
Step 1: multiply by the conjugate of the denominator (\(1 + i\))
\(\frac{3 + 2i}{1 - i} \cdot \frac{1 + i}{1 + i}\)
Step 2: compute the denominator
\((1 - i)(1 + i) = 1^2 + 1^2 = 2\)
Step 3: compute the numerator
\((3 + 2i)(1 + i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i\)
Step 4: divide
\(\frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i\)
📋 Summary Table – Basic Operations
| Operation | Formula |
|---|---|
| Addition | \((a+bi) + (c+di) = (a+c) + (b+d)i\) |
| Subtraction | \((a+bi) - (c+di) = (a-c) + (b-d)i\) |
| Multiplication | \((a+bi)(c+di) = (ac-bd) + (ad+bc)i\) |
| Conjugate | \(\overline{a+bi} = a - bi\) |
| \(z \cdot \bar{z}\) | \(a^2 + b^2\) |
| Division | Multiply numerator and denominator by the conjugate of the denominator |
💡 Tips for the Exam
1️⃣ Powers of i
Period 4: divide by 4 and look at the remainder
2️⃣ Multiplication
Expand brackets normally, remember \(i^2 = -1\)
3️⃣ Division
Always multiply by the conjugate of the denominator!
4️⃣ Equality
Equate real to real, imaginary to imaginary
📝 Summary – Part 1
\(z = a + bi\)
\(i^2 = -1\)
\(\bar{z} = a - bi\)
In the next part: modulus, graphical representation and the complex plane