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.
Today's hook, Try to solve $x^2 + 1 = 0$ using only real numbers. You cannot, every real squares to something $\geq 0$. Now suppose a number $i$ exists with $i^2 = -1$. Without reading on, write down what $(2 + 3i) + (4 + 5i)$ and $(1 + i)^2$ should equal if ordinary algebra still applies. Check after card 05.
0/5QUESTS
1
You are here
Recall, your gut answer first
+5 XP warm-up Without using a calculator, list as many solutions to $x^2 = 9$ as you can in $\mathbb{R}$.
01
Recall, your gut answer first
+5 XP warm-up
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.
auto-saved
2
The two moves for handling a complex number
+5 XP to read 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.
02
The two moves for handling a complex number
+5 XP to read
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 $i^2 = -1$.
Every real $a$ is the complex number $a + 0i$. So $\mathbb{C}$ extends $\mathbb{R}$, it does not replace it. The reals sit on the horizontal axis of the complex plane.
$\operatorname{Im}(z)$ is a real number
If $z = 3 - 7i$ then $\operatorname{Im}(z) = -7$, not $-7i$. The imaginary part is the real coefficient of $i$, without the $i$ attached.
Equality is two equations
$a + bi = c + di$ iff $a = c$ AND $b = d$. One complex equation gives you two real equations to work with, a powerful tool.
03
What you'll master
Know
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 $\operatorname{Im} = 0$
Understand
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
Can do
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
04
Key terms
Imaginary unit ($i$)The symbol $i$ satisfying $i^2 = -1$. It is not a real number; it is a new element adjoined to $\mathbb{R}$ to form $\mathbb{C}$.
Complex number ($z$)An expression $z = a + bi$ with $a, b \in \mathbb{R}$. The set of all complex numbers is denoted $\mathbb{C}$.
Real part ($\operatorname{Re}(z)$)If $z = a + bi$, then $\operatorname{Re}(z) = a$. Always a real number.
Imaginary part ($\operatorname{Im}(z)$)If $z = a + bi$, then $\operatorname{Im}(z) = b$ (a real number, not $bi$).
Purely imaginaryA complex number $z$ with $\operatorname{Re}(z) = 0$ and $\operatorname{Im}(z) \neq 0$. Example: $5i$.
Equality in $\mathbb{C}$$a + bi = c + di \iff a = c \text{ and } b = d$. One complex equation = two real equations.
MEX-N1NESA outcome (Introduction to Complex Numbers): uses the complex number system, including the form $a + bi$, the imaginary unit $i$, and the real and imaginary parts of a complex number.
3
The imaginary unit and standard form
Inside the equation has no solution: every real number squares to a non-negative value.
05
The imaginary unit and standard form
core concept
Inside $\mathbb{R}$ the equation $x^2 + 1 = 0$ has no solution: every real number squares to a non-negative value. To repair this, we postulate a new symbol $i$, the imaginary unit, with the single defining property
$$i^2 = -1.$$
From this single rule the whole of $\mathbb{C}$ is generated. A complex number is any expression
$$z = a + bi, \quad a, b \in \mathbb{R}.$$
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$.
$(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$, expand using ordinary algebra, then replace $i^2$ by $-1$.
Why $\mathbb{R} \subset \mathbb{C}$. If $b = 0$ in $a + bi$, the number is just $a$, a real. So $\mathbb{R}$ embeds inside $\mathbb{C}$ as the set $\{a + 0i : a \in \mathbb{R}\}$. Complex numbers extend the reals; they do not contradict them.
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)$?
4
Equality and square roots of negatives
We just saw that with defines a complex number, with and both real.
06
Equality and square roots of negatives
core concept
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 \qquad (a,b,c,d \in \mathbb{R})$$
Common mistake. The rule $\sqrt{xy} = \sqrt{x}\sqrt{y}$ fails when both $x, y$ are negative. Writing $\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{1} = 1$ gives the wrong answer, the correct value is $i \cdot i = i^2 = -1$. Convert each negative square root to $i$-form before multiplying.
$a + bi = c + di \iff a = c \text{ and } b = d$ · 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 $y = -2$.
5
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
PROBLEM 1 · STANDARD FORM AND PARTS
Write $z = \sqrt{-49} - 6$ in the form $a + bi$ and state $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$.
Convert $\sqrt{-25}$ to $5i$ immediately. Keep both signs to capture both roots.
3
Write in $a + bi$ form: $x = 0 + 5i$ or $x = 0 - 5i$. Both have $\operatorname{Re}(x) = 0$, so both are purely imaginary.
Explicitly showing the $0 + bi$ form makes Re/Im unambiguous and matches the standard form convention.
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).
6
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Trap 01
Treating $\operatorname{Im}(z)$ as having an $i$ in it
If $z = 3 + 5i$, then $\operatorname{Im}(z) = 5$, not $5i$. The imaginary part is a real number, the coefficient of $i$. Writing "$\operatorname{Im}(z) = 5i$" is a marking error.
Trap 02
Using $\sqrt{xy} = \sqrt{x}\sqrt{y}$ with negatives
$\sqrt{-1}\sqrt{-1} \neq \sqrt{(-1)(-1)} = 1$. The correct calculation is $i \cdot i = i^2 = -1$. Convert each negative square root to $i$-form before any multiplication.
Trap 03
Confusing "real" with "imaginary part zero"
A complex number is real when $\operatorname{Im}(z) = 0$ (e.g. $z = 4$). It is purely imaginary when $\operatorname{Re}(z) = 0$ and $\operatorname{Im}(z) \neq 0$ (e.g. $z = 5i$). The number $0$ is technically both.
Did you get this? True or false: the calculation $\sqrt{-4}\sqrt{-9} = \sqrt{(-4)(-9)} = \sqrt{36} = 6$ is correct.
7
Revisit your thinking
Activities · practice with the ideas
Activities · practice with the ideas
1
Express each in standard form $a + bi$ and state $\operatorname{Re}$ and $\operatorname{Im}$: (a) $\sqrt{-16}$, (b) $3 - \sqrt{-50}$, (c) $\sqrt{-1} + 2$.
2
Find real $x$ and $y$ such that $(x + 2) + (3y - 1)i = 7 + 8i$.
3
Solve over $\mathbb{C}$: (a) $x^2 + 4 = 0$, (b) $x^2 + 7 = 0$. Write answers in standard form.
4
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.
5
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?
11
Revisit your thinking
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.
auto-saved
1
Show what you have learned
Multiple choice, then short answer under exam conditions.
01
Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
02
Short answer
ApplyBand 32 marks
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)
auto-saved
ApplyBand 43 marks
Q2. Find all real values of $x$ and $y$ such that $(2x + y) + (x - y)i = 7 - i$. (3 marks)
auto-saved
AnalyseBand 53 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)
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.
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].
1
Consolidate and move on
Sit the module quiz, then close the lesson off.
01
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
