Arithmetic of Complex Numbers
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.
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.
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: $(a+bi)(c+di) = (ac-bd) + (ad+bc)i$
Key facts
- $(a + bi) + (c + di) = (a + c) + (b + d)i$
- $(a + bi) - (c + di) = (a - c) + (b - d)i$
- $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$
- $i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$ (period 4)
Concepts
- Why $\mathbb{C}$ inherits all field axioms (commutative, associative, distributive) from $\mathbb{R}$
- Why complex addition behaves like vector addition
- Why $i^n$ depends only on $n \bmod 4$
Skills
- Add and subtract complex numbers, returning answers in $a + bi$ form
- Multiply complex numbers using FOIL plus $i^2 = -1$
- Reduce $i^n$ for any integer $n$ using the period-4 cycle
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:
- $(3 + 2i) + (1 - 5i) = (3 + 1) + (2 - 5)i = 4 - 3i$.
- $(2 + i)(3 - i) = 6 - 2i + 3i - i^2 = 6 + i - (-1) = 7 + i$.
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 + $i^2 = -1$ 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:
- $i^1 = i$
- $i^2 = -1$
- $i^3 = i^2 \cdot i = -i$
- $i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$
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
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
Did you get this? True or false: $(7 + 3i) - (2 + 5i) = 5 + 8i$.
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.
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)
Comprehensive answers (click to reveal)
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].
Five timed questions on adding, subtracting, multiplying complex numbers, and reducing powers of $i$. 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 arithmetic questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.