Complex numbers and basic algebra

To solve $x^2 + 1 = 0$, we invent a symbol $i$ with $i^2 = -1$, and form new numbers $a + bi$ where $a, b$ are real. These are the complex numbers. Once you accept the single rule $i^2 = -1$, the algebra is identical to the algebra of polynomials in $i$.

Definitions

The imaginary unit is $i$, characterised by $i^2 = -1.$

A complex number is an expression of the form $z = a + bi, \quad a, b \in \mathbb{R}.$

We call $a = \text{Re}(z)$ the real part and $b = \text{Im}(z)$ the imaginary part. (Note: the imaginary part is the real number $b$, not $bi$.) The set of all complex numbers is denoted $\mathbb{C}$.

Equality. $a + bi = c + di \iff a = c$ and $b = d$.

A complex number with $b = 0$ is a real number. A complex number with $a = 0$ and $b \ne 0$ is purely imaginary.

The four arithmetic operations

For $z_1 = a + bi$ and $z_2 = c + di$:

Addition: $z_1 + z_2 = (a + c) + (b + d)i$.

Subtraction: $z_1 - z_2 = (a - c) + (b - d)i$.

Multiplication: distribute and use $i^2 = -1$: $(a + bi)(c + di) = ac + adi + bci + bd \cdot i^2 = (ac - bd) + (ad + bc)i.$

Division: multiply numerator and denominator by the conjugate of the denominator. To compute $\dfrac{a + bi}{c + di}$ (with $c + di \ne 0$): $\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}.$

Powers of $i$

$i^1 = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1,\quad i^5 = i,\ \dots$

The pattern has period $4$. In general $i^n = i^{n \bmod 4}$.

Algebraic structure

$\mathbb{C}$ is a field: addition and multiplication are commutative, associative, distributive, and every non-zero element has a multiplicative inverse. So all the algebraic moves you know , collecting like terms, factoring, distributing , work in $\mathbb{C}$ just as in $\mathbb{R}$.

One difference: $\mathbb{C}$ is not ordered. There is no notion of "$z_1 < z_2$" between complex numbers. The relations $<, >, \le, \ge$ apply only to real numbers.

Worked examples

Example 1. Simplify $(3 + 2i) + (5 - 4i)$.

$(3 + 5) + (2 - 4)i = 8 - 2i$.

Example 2. Compute $(2 + 3i)(1 - 4i)$.

$2 \cdot 1 + 2 \cdot (-4i) + 3i \cdot 1 + 3i \cdot (-4i) = 2 - 8i + 3i + 12 = 14 - 5i$.

Example 3. Find $\dfrac{1 + i}{1 - i}$.

Multiply by $\dfrac{1 + i}{1 + i}$: $\frac{(1 + i)^2}{(1 - i)(1 + i)} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i.$

Example 4. Compute $i^{2025}$.

$2025 = 4 \cdot 506 + 1$, so $i^{2025} = i^1 = i$.

Example 5 (harder). Find real numbers $a, b$ such that $(a + bi)^2 = 5 - 12i$.

Expanding: $a^2 - b^2 + 2abi = 5 - 12i$. So $a^2 - b^2 = 5$ and $2ab = -12$, giving $ab = -6$.

From $ab = -6$: $b = -6/a$. Substitute: $a^2 - 36/a^2 = 5$, so $a^4 - 5a^2 - 36 = 0$. Let $u = a^2$: $u^2 - 5u - 36 = 0$, $u = \dfrac{5 \pm \sqrt{25 + 144}}{2} = \dfrac{5 \pm 13}{2}$. So $u = 9$ or $u = -4$. Since $a$ is real, $u = 9$, $a = \pm 3$.

If $a = 3$: $b = -2$. If $a = -3$: $b = 2$. So $a + bi = 3 - 2i$ or $-3 + 2i$.

Try it yourself

1. Simplify $(4 + 5i) - (2 - 3i)$.
2. Multiply $(1 + 2i)(3 + 4i)$.
3. Find $i^{50}$.
4. Compute $\dfrac{2 - 3i}{1 + i}$.
5. Solve $z + (3 + 2i) = 5 - i$ for $z$.
6. Find real numbers $x, y$ such that $(x + iy)(2 + 3i) = 8 + i$.
7. Compute $(1 + i)^4$.
8. Find $\dfrac{1}{2 + 3i}$.
9. Express $\dfrac{(1 - i)^2}{1 + i}$ in $a + bi$ form.
10. Compute $i + i^2 + i^3 + \dots + i^{100}$.
11. If $z = 3 + 4i$, compute $z^2$, $\dfrac{1}{z}$, $z \cdot \dfrac{1}{z}$.
12. Find $\sqrt{i}$ (in $a + bi$ form).

Pitfalls / Tricks

• $i = \sqrt{-1}$ is a symbol, not a real number. Many calculator-style mistakes come from treating $i$ as if it were $\pm 1$ in some clever way.
• The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ fails if both $a, b$ are negative. E.g. $\sqrt{-2}\sqrt{-3} \ne \sqrt{6}$; it equals $-\sqrt{6}$.
• $\mathbb{C}$ is not ordered: writing "$2 + 3i > 1$" is meaningless.
• Insight. Every complex computation reduces to: distribute, use $i^2 = -1$, collect real and imaginary parts. There is no other technique.