Once the symbol $i$ obeys $i^2 = -1$, complex numbers add, subtract and multiply exactly like algebraic binomials in $i$, collect like terms, expand brackets, replace $i^2$ with $-1$ at the end. The only new pattern is the cyclic behaviour of powers of $i$: every fourth power returns to $1$, so any $i^n$ reduces to one of four values.
+5 XP warm-up Expand $(a + b)(c + d)$ with real $a, b, c, d$.
Expand $(a + b)(c + d)$ with real $a, b, c, d$. Now expand $(a + bi)(c + di)$, same FOIL pattern, then collect the $i^2$ term. Before checking what does the product equal in $a + bi$ form? Sketch the expansion below.
+5 XP to read Every complex computation rewards two reflexes: treat like an algebraic variable distribute, expand, collect,...
Every complex computation rewards two reflexes: treat $i$ like an algebraic variable distribute, expand, collect, then substitute $i^2 = -1$ at the very last moment and re-collect into $a + bi$ form. For powers of $i$, use the cycle of four: $i^1, i^2, i^3, i^4 = i, -1, -i, 1$.
The expand-replace-collect reading: (1) expand brackets treating $i$ as a variable, (2) replace every $i^2$ with $-1$ (and $i^3$ with $-i$, $i^4$ with $1$), (3) collect like terms into $a + bi$.
Sum: $(a+bi) + (c+di) = (a+c) + (b+d)i$ · Product: $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$
Because satisfies the same field axioms as , you manipulate exactly like an algebraic binomial in the "variable" , then...
Because $\mathbb{C}$ satisfies the same field axioms as $\mathbb{R}$, you manipulate $a + bi$ exactly like an algebraic binomial in the "variable" $i$, then replace $i^2$ by $-1$ when it appears.
Addition / subtraction are componentwise:
Multiplication uses FOIL plus the identity $i^2 = -1$:
Worked through the hook:
Add / subtract componentwise: real with real, imaginary with imaginary · Multiply with FOIL, then replace $i^2$ with $-1$ · $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$, memorise or re-derive each time · Always finish in $a + bi$ form (real coefficient first, then $i$-term)
Pause, copy componentwise addition, the FOIL multiplication formula $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$, and the rule to always finish in $a+bi$ form into your book.
Quick check: Compute $(4 + 3i)(2 - i)$ in $a + bi$ form.
We just saw that addition, subtraction and multiplication of complex numbers all act componentwise (with FOIL + for...
We just saw that addition, subtraction and multiplication of complex numbers all act componentwise (with FOIL + $i^2 = -1$ for multiplication), giving the formula $(a+bi)(c+di) = (ac-bd)+(ad+bc)i$. That raises a question: what happens when you raise $i$ to large integer powers? This card answers it → $i^n$ has period 4: the cycle $i, -1, -i, 1$ repeats, so compute via $n \bmod 4$.
From $i^2 = -1$ the higher powers fall out instantly:
From $i^4 = 1$ the pattern repeats: $i^5 = i$, $i^6 = -1$, and so on. So $i^n$ depends only on the remainder of $n$ when divided by 4:
Worked example from the hook: $i^{2026}$. Compute $2026 \div 4 = 506$ remainder $2$. So $i^{2026} = i^2 = -1$.
$i^1, i^2, i^3, i^4 = i, -1, -i, 1$, period 4 · $i^n$ via $n \bmod 4$: remainder 0, 1, 2, 3 gives $1, i, -1, -i$ · Worked: $i^{2026}$, $2026 = 4 \cdot 506 + 2$, so $i^{2026} = i^2 = -1$ · $i^{-1} = -i$, and the cycle extends to negative exponents
Pause, copy the period-4 cycle $i^1=i,\, i^2=-1,\, i^3=-i,\, i^4=1$, the reduction $i^n$ via $n \bmod 4$, and the worked example $i^{2026} = i^2 = -1$ into your book.
Did you get this? True or false: $i^{99} = -i$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Let $z_1 = 5 - 2i$ and $z_2 = -3 + 4i$. Compute $z_1 + z_2$ and $z_1 - z_2$, leaving each answer in $a + bi$ form.
Compute $(3 - 2i)(4 + 5i)$ in $a + bi$ form.
Evaluate (a) $i^{17}$, (b) $i^{50} - i^{75}$, and (c) $(1 + i)^2$.
Fill the gap: $(a + bi)(c + di) = ($ $) + ($ $)i$, derived by FOIL and using $i^2 = -1$.
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $(7 + 3i) - (2 + 5i) = 5 + 8i$.
Activities · practice with the ideas
Activities · practice with the ideas
Compute, leaving answers in $a + bi$ form: (a) $(2 + 7i) + (5 - 3i)$, (b) $(6 - i) - (4 + 2i)$, (c) $3(1 - 2i) + 4(2 + i)$.
Multiply, leaving answers in $a + bi$ form: (a) $(1 + i)(2 + 3i)$, (b) $(4 - i)(4 + i)$, (c) $(5i)(2 - 3i)$.
Evaluate: (a) $i^{23}$, (b) $i^{100}$, (c) $i^{-7}$.
Expand $(2 + 3i)^2$ and $(1 - i)^3$. Leave each in $a + bi$ form.
Find real $x$ and $y$ such that $(x + iy)(2 + i) = 5 - 5i$. (Hint: expand the product first, then equate components.)
Odd one out: Three of these expressions equal $i$. Which one does NOT?
Earlier you computed $(3 + 2i) + (1 - 5i)$, $(2 + i)(3 - i)$ and $i^{2026}$ before any formal rules were stated.
The first ($4 - 3i$) is componentwise addition; the second ($7 + i$) is FOIL plus $i^2 = -1$; the third ($-1$) is the period-4 cycle. Notice that the second result $7 + i$ stays in $a + bi$ form because the $-i^2$ from the FOIL becomes a $+1$ joining the real part. This idea, that "multiplying two complex numbers produces a complex number", is what makes $\mathbb{C}$ a field, closed under arithmetic, and is the foundation for division and modulus in the next lessons.
Multiple choice, then short answer under exam conditions.
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. Let $z_1 = 4 - 3i$ and $z_2 = -1 + 5i$. Compute $z_1 + 2z_2$ in $a + bi$ form. (2 marks)
Q2. Expand $(3 + 2i)(1 - 4i)$ in $a + bi$ form, showing the $i^2 = -1$ step explicitly. (3 marks)
Q3. (a) Simplify $i^{2026} + i^{2027} + i^{2028} + i^{2029}$. (b) Explain why $i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$ for every integer $n$. (3 marks)
Activity answers:
1. (a) $7 + 4i$. (b) $(6 - 4) + (-1 - 2)i = 2 - 3i$. (c) $3(1 - 2i) + 4(2 + i) = (3 - 6i) + (8 + 4i) = 11 - 2i$.
2. (a) $(1 + i)(2 + 3i) = 2 + 3i + 2i + 3i^2 = 2 + 5i - 3 = -1 + 5i$. (b) $(4 - i)(4 + i) = 16 - i^2 = 16 + 1 = 17$. (c) $(5i)(2 - 3i) = 10i - 15i^2 = 15 + 10i$.
3. (a) $i^{23} = i^{20} \cdot i^3 = 1 \cdot (-i) = -i$. (b) $i^{100} = (i^4)^{25} = 1$. (c) $i^{-7} = 1/i^7 = 1/(-i) = i$ (multiply numerator and denominator by $i$).
4. $(2 + 3i)^2 = 4 + 12i + 9i^2 = -5 + 12i$. $(1 - i)^2 = 1 - 2i + i^2 = -2i$, so $(1 - i)^3 = (1 - i)(-2i) = -2i + 2i^2 = -2 - 2i$.
5. $(x + iy)(2 + i) = (2x - y) + (x + 2y)i = 5 - 5i$. So $2x - y = 5$ and $x + 2y = -5$. Solve: from the first $y = 2x - 5$; substitute: $x + 2(2x - 5) = -5 \Rightarrow 5x = 5 \Rightarrow x = 1$, then $y = -3$.
Q1 (2 marks): $2z_2 = -2 + 10i$ [1]. $z_1 + 2z_2 = (4 - 2) + (-3 + 10)i = 2 + 7i$ [1].
Q2 (3 marks): FOIL: $(3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2$ [1]. Replace $i^2 = -1$: $-8i^2 = +8$, giving $3 - 12i + 2i + 8$ [1]. Collect: $(3 + 8) + (-12 + 2)i = 11 - 10i$ [1].
Q3 (3 marks): (a) $2026 \equiv 2 \pmod 4$, so $i^{2026} = -1$, $i^{2027} = -i$, $i^{2028} = 1$, $i^{2029} = i$. Sum $= -1 - i + 1 + i = 0$ [1]. (b) Factor $i^n$: $i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i + i^2 + i^3)$ [1]. The bracket equals $1 + i - 1 - i = 0$, so the whole sum is $0$ for every integer $n$ [1].
Sit the module quiz, then close the lesson off.
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.
Start the module quiz →Tick when you've finished the practice and review.
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