Complex Number Arithmetic

With the imaginary unit defined, we now do algebra with complex numbers: add, subtract, multiply and divide them, always finishing in the standard form $a + bi$a+bi. The single new rule $i^2 = -1$i2=−1 is all that changes; everything else is the bracket-expansion and surd-rationalising you already know. These four operations underpin every later topic — Argand geometry, polar form, De Moivre — so fluency here pays off repeatedly. They also matter far beyond the exam: complex arithmetic is the native language of alternating-current circuit analysis, signal processing and quantum mechanics, where dividing by a complex impedance or amplitude is an everyday calculation. Mastering "realise the denominator" now is therefore an investment well past A-Level.

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Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2. It is overwhelmingly AO1 (carry out routine arithmetic procedures accurately), with AO2 appearing when you justify why dividing by the conjugate works (it produces a real denominator). The "realising the denominator" technique reappears constantly — in modulus–argument conversions, in solving equations with complex coefficients, and in roots-of-unity calculations — so treat it as a core skill, not a one-off trick.

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Core theory: the four operations

Addition and subtraction

Combine the real and imaginary parts separately — this is exactly vector addition in disguise (see Argand diagrams):

$(a + bi) + (c + di) = (a + c) + (b + d)i,$(a+bi)+(c+di)=(a+c)+(b+d)i, $(a + bi) - (c + di) = (a - c) + (b - d)i.$(a+bi)−(c+di)=(a−c)+(b−d)i.

These are the easy operations: no $i^2$i2 appears, so there are no sign traps. The only care needed is with subtraction, where the minus sign must be distributed across both parts of the second number — $(7 - i) - (4 + 3i) = (7 - 4) + (-1 - 3)i = 3 - 4i$(7−i)−(4+3i)=(7−4)+(−1−3)i=3−4i, not $3 + 2i$3+2i. A quick reality check: $\operatorname{Re}$Re and $\operatorname{Im}$Im are themselves additive, meaning $\operatorname{Re}(z + w) = \operatorname{Re}(z) + \operatorname{Re}(w)$Re(z+w)=Re(z)+Re(w) and similarly for $\operatorname{Im}$Im, which is just a restatement of the rule above.

Multiplication

Expand by the distributive law (FOIL), then replace $i^2$i2 by $-1$−1 and collect:

$\begin{aligned} (a + bi)(c + di) &= ac + adi + bci + bd\,i^2 \\ &= ac + adi + bci - bd \\ &= (ac - bd) + (ad + bc)i. \end{aligned}$(a+bi)(c+di)​=ac+adi+bci+bdi2=ac+adi+bci−bd=(ac−bd)+(ad+bc)i.​

You do not need to memorise this formula — just expand and simplify each time, which is less error-prone. Multiplication by a real scalar $k$k simply scales both parts: $k(a + bi) = ka + kbi$k(a+bi)=ka+kbi. Notice the structure of the answer: the real part $ac - bd$ac−bd has a minus (because the $bi \cdot di$bi⋅di term carries $i^2 = -1$i2=−1), while the imaginary part $ad + bc$ad+bc is a plus. Getting that sign pattern right is the single most important habit in complex multiplication. Two special cases recur constantly:

$(a + bi)^2 = a^2 - b^2 + 2abi, \qquad (a + bi)(a - bi) = a^2 + b^2.$(a+bi)2=a2−b2+2abi,(a+bi)(a−bi)=a2+b2.

The first is the square (note the $a^2 - b^2$a2−b2 real part and $2ab$2ab imaginary part); the second is the conjugate product, which is the entire basis of division below.

The conjugate product — the key to division

The complex conjugate of $c + di$c+di is $c - di$c−di (sign of the imaginary part flipped); it is the workhorse of complex division, just as the surd conjugate is the workhorse of surd rationalisation. (The conjugate gets a full treatment, including the conjugate root theorem, in the next lesson — here we need only its product property.) Their product is always real and non-negative:

$(c + di)(c - di) = c^2 - (di)^2 = c^2 - d^2 i^2 = c^2 + d^2.$(c+di)(c−di)=c2−(di)2=c2−d2i2=c2+d2.

This is the engine of division. To divide, multiply numerator and denominator by the conjugate of the denominator — the complex analogue of rationalising a surd:

$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}.$c+dia+bi​=(c+di)(c−di)(a+bi)(c−di)​=c2+d2(a+bi)(c−di)​.

Because the denominator is now the real number $c^2 + d^2$c2+d2, the quotient drops straight into the form $a + bi$a+bi. Carrying the general expansion through once shows exactly what comes out:

$\frac{a + bi}{c + di} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + \frac{bc - ad}{c^2 + d^2}\,i.$c+dia+bi​=c2+d2(ac+bd)+(bc−ad)i​=c2+d2ac+bd​+c2+d2bc−ad​i.

You should not memorise this — re-deriving it each time by multiplying by the conjugate is far safer — but it confirms the answer is always a genuine complex number with real and imaginary parts. This process is called realising (or rationalising) the denominator, and it is the exact analogue of rationalising $\tfrac{1}{c + d\sqrt 2}$c+d2​1​ by multiplying by $c - d\sqrt 2$c−d2​.

Powers by repeated multiplication

To raise a complex number to a small integer power, the safest route is repeated squaring/multiplying rather than the binomial theorem. For $z^3$z3, compute $z^2$z2 first and then multiply by $z$z (Example 3). For $z^4$z4, compute $z^2$z2 and square it: $z^4 = (z^2)^2$z4=(z2)2. This keeps each step to a product of two complex numbers — minimal terms, minimal slips. (For large powers you will soon prefer polar form and De Moivre's theorem, but for $z^2, z^3, z^4$z2,z3,z4 direct multiplication is quick and reliable.)

Algebraic structure

Complex arithmetic obeys every familiar field law: addition and multiplication are commutative and associative, multiplication distributes over addition, $0$0 and $1$1 are the identities, and every non-zero $z$z has an inverse $1/z$1/z. The zero-product property also holds: if $z_1 z_2 = 0$z1​z2​=0 then $z_1 = 0$z1​=0 or $z_2 = 0$z2​=0 (proof: if $z_1 \neq 0$z1​=0, multiply by $z_1^{-1}$z1−1​ to force $z_2 = 0$z2​=0). In short, $\mathbb{C}$C is a field — you may manipulate complex expressions with exactly the same confidence as real ones, provided you remember $i^2 = -1$i2=−1. One thing you lose compared with $\mathbb{R}$R: there is no order, so it is meaningless to call one complex number "bigger" than another, and inequalities only ever apply to real quantities such as moduli $|z|$∣z∣.

Law	Statement
Commutativity	$z_1 + z_2 = z_2 + z_1$z1​+z2​=z2​+z1​, $\;z_1 z_2 = z_2 z_1$z1​z2​=z2​z1​
Associativity	$(z_1 z_2)z_3 = z_1(z_2 z_3)$(z1​z2​)z3​=z1​(z2​z3​)
Distributivity	$z_1(z_2 + z_3) = z_1 z_2 + z_1 z_3$z1​(z2​+z3​)=z1​z2​+z1​z3​
Identities	$z + 0 = z$z+0=z, $\;z \cdot 1 = z$z⋅1=z
Conjugate product	$(a+bi)(a-bi) = a^2 + b^2 \in \mathbb{R}_{\geq 0}$(a+bi)(a−bi)=a2+b2∈R≥0​

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Worked examples (with mark scheme)

Example 0 — combined operations

Given $z = 5 + 2i$z=5+2i and $w = -3 + 7i$w=−3+7i, find $z + w$z+w, $z - w$z−w and $3z - 2w$3z−2w.

$z + w = (5 - 3) + (2 + 7)i = 2 + 9i; \quad z - w = (5 + 3) + (2 - 7)i = 8 - 5i.$z+w=(5−3)+(2+7)i=2+9i;z−w=(5+3)+(2−7)i=8−5i. $3z - 2w = (15 + 6i) - (-6 + 14i) = (15 + 6) + (6 - 14)i = 21 - 8i.$3z−2w=(15+6i)−(−6+14i)=(15+6)+(6−14)i=21−8i.

(B1 each. Note how $3z - 2w$3z−2w scales each number first, then subtracts componentwise — the distributive and additive rules working together.)

Example 1 — multiplication

Compute $(2 + 3i)(4 - i)$(2+3i)(4−i).

$\begin{aligned} (2 + 3i)(4 - i) &= 8 - 2i + 12i - 3i^2 \quad &(\text{M1 expand all four products}) \\ &= 8 + 10i - 3(-1) \quad &(\text{M1 use } i^2 = -1) \\ &= 11 + 10i. \quad &(\text{A1}) \end{aligned}$(2+3i)(4−i)​=8−2i+12i−3i2=8+10i−3(−1)=11+10i.​(M1 expand all four products)(M1 use i2=−1)(A1)​

(M1 expansion; M1 substituting $i^2 = -1$i2=−1; A1 correct $a + bi$a+bi form.)

Example 2 — division (realising the denominator)

Express $\dfrac{3 + i}{1 - 2i}$1−2i3+i​ in the form $a + bi$a+bi.

$\begin{aligned} \frac{3 + i}{1 - 2i} &= \frac{(3 + i)(1 + 2i)}{(1 - 2i)(1 + 2i)} \quad &(\text{M1 multiply by conjugate } 1+2i) \\ &= \frac{3 + 6i + i + 2i^2}{1 + 4} = \frac{1 + 7i}{5} \quad &(\text{M1 expand, simplify denominator}) \\ &= \frac{1}{5} + \frac{7}{5}i. \quad &(\text{A1}) \end{aligned}$1−2i3+i​​=(1−2i)(1+2i)(3+i)(1+2i)​=1+43+6i+i+2i2​=51+7i​=51​+57​i.​(M1 multiply by conjugate 1+2i)(M1 expand, simplify denominator)(A1)​

(M1 choosing the correct conjugate; M1 expanding both products; A1 final form. Note the denominator $(1-2i)(1+2i)=1+4=5$(1−2i)(1+2i)=1+4=5 is real, as it must be.)

Example 3 — a higher power by repeated multiplication

If $z = 1 + 2i$z=1+2i, find $z^3$z3.

$z^2 = (1 + 2i)^2 = 1 + 4i + 4i^2 = -3 + 4i. \quad (\text{M1})$z2=(1+2i)2=1+4i+4i2=−3+4i.(M1) $z^3 = z^2 \cdot z = (-3 + 4i)(1 + 2i) = -3 - 6i + 4i + 8i^2 = -11 - 2i. \quad (\text{M1 A1})$z3=z2⋅z=(−3+4i)(1+2i)=−3−6i+4i+8i2=−11−2i.(M1 A1)

(M1 for $z^2$z2; M1 for multiplying by $z$z; A1 for $-11 - 2i$−11−2i.)

Example 4 — solving a linear equation with complex coefficients

Solve $(1 + i)z = 3 - i$(1+i)z=3−i for $z$z, giving your answer in the form $a + bi$a+bi.

Make $z$z the subject and realise the denominator:

$z = \frac{3 - i}{1 + i} = \frac{(3 - i)(1 - i)}{(1 + i)(1 - i)} \quad (\text{M1 divide, multiply by conjugate})$z=1+i3−i​=(1+i)(1−i)(3−i)(1−i)​(M1 divide, multiply by conjugate) $= \frac{3 - 3i - i + i^2}{1 + 1} = \frac{2 - 4i}{2} = 1 - 2i. \quad (\text{M1 expand; A1})$=1+13−3i−i+i2​=22−4i​=1−2i.(M1 expand; A1)

(M1 for isolating $z$z as a quotient; M1 for the conjugate multiplication and expansion; A1 for $1 - 2i$1−2i. Check: $(1+i)(1-2i) = 1 - 2i + i - 2i^2 = 3 - i$(1+i)(1−2i)=1−2i+i−2i2=3−i ✓.)

Example 5 — equating parts after a product

Find real numbers $a$a and $b$b such that $(a + bi)(2 - i) = 7 + i$(a+bi)(2−i)=7+i.

Expand the left-hand side:

$(a + bi)(2 - i) = 2a - ai + 2bi - bi^2 = (2a + b) + (2b - a)i. \quad (\text{M1 expand and collect})$(a+bi)(2−i)=2a−ai+2bi−bi2=(2a+b)+(2b−a)i.(M1 expand and collect)

Compare with $7 + i$7+i:

$2a + b = 7, \qquad 2b - a = 1. \quad (\text{M1 form two real equations})$2a+b=7,2b−a=1.(M1 form two real equations)

From the first, $b = 7 - 2a$b=7−2a; substituting, $2(7 - 2a) - a = 1 \Rightarrow 14 - 5a = 1 \Rightarrow a = \tfrac{13}{5}$2(7−2a)−a=1⇒14−5a=1⇒a=513​, and $b = 7 - \tfrac{26}{5} = \tfrac{9}{5}$b=7−526​=59​. (A1 both values.) (Alternatively, divide: $a + bi = \tfrac{7 + i}{2 - i} = \tfrac{(7+i)(2+i)}{5} = \tfrac{13 + 9i}{5}$a+bi=2−i7+i​=5(7+i)(2+i)​=513+9i​, giving the same $a = \tfrac{13}{5}, b = \tfrac{9}{5}$a=513​,b=59​ — a useful cross-check.)

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Specimen-style exam question

(specimen-style — not from any past paper)

Given $z = 2 + i$z=2+i and $w = 3 - 2i$w=3−2i, find, in the form $a + bi$a+bi: (a) $zw$zw; (b) $\dfrac{z}{w}$wz​; (c) $z^2 - w$z2−w.

(a) $zw = (2 + i)(3 - 2i) = 6 - 4i + 3i - 2i^2 = 8 - i.$zw=(2+i)(3−2i)=6−4i+3i−2i2=8−i.

(b) $\dfrac{z}{w} = \dfrac{(2 + i)(3 + 2i)}{(3 - 2i)(3 + 2i)} = \dfrac{6 + 4i + 3i + 2i^2}{9 + 4} = \dfrac{4 + 7i}{13} = \dfrac{4}{13} + \dfrac{7}{13}i.$wz​=(3−2i)(3+2i)(2+i)(3+2i)​=9+46+4i+3i+2i2​=134+7i​=134​+137​i.

(c) $z^2 = (2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$z2=(2+i)2=4+4i+i2=3+4i, so $z^2 - w = (3 + 4i) - (3 - 2i) = 6i.$z2−w=(3+4i)−(3−2i)=6i.

(d) Hence find $z^2 + w^2$z2+w2.

Reusing $z^2 = 3 + 4i$z2=3+4i and computing $w^2 = (3 - 2i)^2 = 9 - 12i + 4i^2 = 5 - 12i$w2=(3−2i)2=9−12i+4i2=5−12i,

$z^2 + w^2 = (3 + 4i) + (5 - 12i) = 8 - 8i.$z2+w2=(3+4i)+(5−12i)=8−8i.