Introduction to Complex Numbers
The equation $x^2 = -1$ has no solution in the reals — yet declaring a single new number $i$ with $i^2 = -1$ produces an entire arithmetic system, the complex numbers $\mathbb{C}$, that extends $\mathbb{R}$ while preserving every algebraic rule you already know. This lesson defines $i$, fixes the standard form $a + bi$, and pins down what it means for two complex numbers to be equal.
Without using a calculator, list as many solutions to $x^2 = 9$ as you can in $\mathbb{R}$. Now list solutions to $x^2 = -9$ in $\mathbb{R}$. Before checking — what is the smallest extension of the number system that would let you solve both? Sketch your thinking below.
Every complex number is treated with two reflexes: write it in standard form $a + bi$ with $a, b \in \mathbb{R}$ — never leave it as $\sqrt{-9}$ — then read off Re and Im. From that point, complex numbers obey the same field rules as reals: associative, commutative, distributive. The only new fact you need is $i^2 = -1$.
The standardise-then-identify reading: (1) rewrite any expression as $a + bi$, (2) name $a = \operatorname{Re}(z)$ and $b = \operatorname{Im}(z)$, (3) treat $i$ as a symbol obeying $i^2 = -1$.
$z = a + bi$ · $\operatorname{Re}(z) = a$ · $\operatorname{Im}(z) = b$ · $i^2 = -1$
Key facts
- $i$ is defined by $i^2 = -1$ (equivalently $i = \sqrt{-1}$)
- A complex number has the form $z = a + bi$ with $a, b \in \mathbb{R}$
- $\operatorname{Re}(z) = a$ and $\operatorname{Im}(z) = b$ (both are real)
- $\mathbb{R} \subset \mathbb{C}$: every real is a complex with $\operatorname{Im} = 0$
Concepts
- Why $\mathbb{C}$ is needed: $x^2 + 1 = 0$ has no real solution
- Why "purely imaginary" means $\operatorname{Re}(z) = 0$, not $\operatorname{Im}(z) = 0$
- Why $a + bi = c + di$ forces both real parts and imaginary parts to match
Skills
- Rewrite $\sqrt{-k}$ ($k > 0$) as $\sqrt{k}\,i$ in standard form
- State $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ from any $a + bi$ expression
- Solve equations of the form $a + bi = c + di$ by equating components
Inside $\mathbb{R}$ the equation $x^2 + 1 = 0$ has no solution: every real number squares to a non-negative value. To repair this, we postulate a new symbol $i$ — the imaginary unit — with the single defining property
From this single rule the whole of $\mathbb{C}$ is generated. A complex number is any expression
The real number $a$ is the real part, written $\operatorname{Re}(z)$; the real number $b$ is the imaginary part, written $\operatorname{Im}(z)$. Notice $\operatorname{Im}(z)$ does not include the $i$ — it is the real coefficient sitting in front of $i$.
Worked through the hook:
- $(2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i$ — collect real and imaginary parts.
- $(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$ — expand using ordinary algebra, then replace $i^2$ by $-1$.
Definition: $i^2 = -1$ (so $i = \sqrt{-1}$) · Standard form: $z = a + bi$, $a, b \in \mathbb{R}$ · $\operatorname{Re}(z) = a$, $\operatorname{Im}(z) = b$ — both are real · Reals sit inside $\mathbb{C}$ as numbers with $\operatorname{Im}(z) = 0$
Pause — copy $i^2 = -1$, the standard form $z = a+bi$, and the definitions $\operatorname{Re}(z) = a$, $\operatorname{Im}(z) = b$ into your book.
Quick check: Let $z = 4 - 7i$. What are $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$?
We just saw that $z = a + bi$ with $i^2 = -1$ defines a complex number, with $\operatorname{Re}(z) = a$ and $\operatorname{Im}(z) = b$ both real. That raises a question: when are two complex numbers equal, and how do we find square roots of negative numbers? This card answers it → equality requires both real and imaginary parts to match, giving two real equations from one complex equation.
Because a complex number carries two real coordinates, equality is stricter than in $\mathbb{R}$: it forces both components to match.
- Equality. $a + bi = c + di \iff a = c \text{ and } b = d$.
- This is enormously useful: one complex equation in $z$ encodes two simultaneous real equations.
Square roots of negative reals. For any $k > 0$, $\sqrt{-k} = \sqrt{k}\;i$. So $\sqrt{-9} = 3i$, $\sqrt{-2} = \sqrt{2}\,i$. Always extract the $i$ — never leave a $\sqrt{-\text{number}}$ in an answer.
$a + bi = c + di \iff a = c \text{ and } b = d$ · One complex equation = two real equations (equate Re and Im) · $\sqrt{-k} = \sqrt{k}\,i$ for $k > 0$ · Trap: $\sqrt{x}\sqrt{y} \neq \sqrt{xy}$ when both $x, y < 0$
Pause — copy the equality rule $a+bi = c+di \iff a=c$ and $b=d$, the identity $\sqrt{-k} = \sqrt{k}\,i$ for $k>0$, and the trap $\sqrt{x}\sqrt{y} \neq \sqrt{xy}$ when both negative into your book.
Did you get this? True or false: if $x + yi = 5 - 2i$ with $x, y \in \mathbb{R}$, then $x = 5$ and $y = -2$.
Worked examples · 3 in a row, reveal as you go
Write $z = \sqrt{-49} - 6$ in the form $a + bi$ and state $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$.
Find real numbers $x$ and $y$ such that $(2x - 1) + (y + 3)i = 5 - 2i$.
Solve $x^2 + 25 = 0$ over $\mathbb{C}$, expressing solutions in the form $a + bi$.
Fill the gap: The defining property of the imaginary unit is $i^2 = $ , and for any real $k > 0$, $\sqrt{-k} = $ (write as √k followed by i).
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the calculation $\sqrt{-4}\sqrt{-9} = \sqrt{(-4)(-9)} = \sqrt{36} = 6$ is correct.
Activities · practice with the ideas
Express each in standard form $a + bi$ and state $\operatorname{Re}$ and $\operatorname{Im}$: (a) $\sqrt{-16}$, (b) $3 - \sqrt{-50}$, (c) $\sqrt{-1} + 2$.
Find real $x$ and $y$ such that $(x + 2) + (3y - 1)i = 7 + 8i$.
Solve over $\mathbb{C}$: (a) $x^2 + 4 = 0$, (b) $x^2 + 7 = 0$. Write answers in standard form.
Decide whether each is true or false. Justify briefly. (a) Every real number is a complex number. (b) $\operatorname{Im}(z)$ is always non-negative. (c) The number $0$ is both real and purely imaginary.
A complex number $z$ satisfies $2z = (x + 6) + (y - 4)i$ where $z = 3 - i$. Find real $x$ and $y$.
Odd one out: Three of these expressions equal $-9$. Which one does NOT?
Earlier you guessed $(2 + 3i) + (4 + 5i)$ and $(1 + i)^2$ before any rules were stated.
The first is a clean component sum: $6 + 8i$. The second, $1 + 2i + i^2 = 2i$, hides a surprise — squaring a non-zero complex number can produce a purely imaginary result. That is impossible in $\mathbb{R}$: no real squares to something off the real axis. This single observation is the entire reason $\mathbb{C}$ is needed and is the gateway to everything in Module 12.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write $z = 5 - \sqrt{-72}$ in the form $a + bi$ with $a, b \in \mathbb{R}$. State $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$. (2 marks)
Q2. Find all real values of $x$ and $y$ such that $(2x + y) + (x - y)i = 7 - i$. (3 marks)
Q3. (a) Solve $2x^2 + 18 = 0$ over $\mathbb{C}$. (b) Explain why the equation has no real solutions but two complex solutions, and classify the solutions as real, purely imaginary, or neither. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. (a) $\sqrt{-16} = 4i = 0 + 4i$; $\operatorname{Re} = 0, \operatorname{Im} = 4$. (b) $3 - \sqrt{-50} = 3 - \sqrt{50}\,i = 3 - 5\sqrt{2}\,i$; $\operatorname{Re} = 3, \operatorname{Im} = -5\sqrt{2}$. (c) $\sqrt{-1} + 2 = i + 2 = 2 + i$; $\operatorname{Re} = 2, \operatorname{Im} = 1$.
2. $x + 2 = 7 \Rightarrow x = 5$; $3y - 1 = 8 \Rightarrow y = 3$.
3. (a) $x^2 = -4 \Rightarrow x = \pm 2i$. (b) $x^2 = -7 \Rightarrow x = \pm\sqrt{7}\,i$. Both pairs are purely imaginary.
4. (a) True — every real $a$ equals $a + 0i \in \mathbb{C}$. (b) False — e.g. $z = 1 - 3i$ has $\operatorname{Im}(z) = -3$. (c) The number $0 = 0 + 0i$ is real ($\operatorname{Im} = 0$); strict definitions exclude it from "purely imaginary" because $\operatorname{Im} \neq 0$ is required.
5. $2z = 2(3 - i) = 6 - 2i$. Equate: $x + 6 = 6 \Rightarrow x = 0$; $y - 4 = -2 \Rightarrow y = 2$.
Q1 (2 marks): $\sqrt{-72} = \sqrt{72}\,i = 6\sqrt{2}\,i$ [1]. So $z = 5 - 6\sqrt{2}\,i$; $\operatorname{Re}(z) = 5$, $\operatorname{Im}(z) = -6\sqrt{2}$ [1].
Q2 (3 marks): Equate real parts: $2x + y = 7$ [1]. Equate imaginary parts: $x - y = -1$ [1]. Adding: $3x = 6 \Rightarrow x = 2$; then $y = 7 - 4 = 3$. So $x = 2, y = 3$ [1].
Q3 (3 marks): (a) $2x^2 + 18 = 0 \Rightarrow x^2 = -9 \Rightarrow x = \pm 3i$ [1]. (b) No real solution because $x^2 \geq 0$ for every real $x$, but the equation requires $x^2 = -9 < 0$ [1]. In $\mathbb{C}$, $i$ satisfies $i^2 = -1$, so $(\pm 3i)^2 = 9 i^2 = -9$ ✓. Both solutions have $\operatorname{Re} = 0$, so they are purely imaginary [1].
Five timed questions on standard form, real and imaginary parts, equality, and square roots of negatives. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick complex-number identification questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.