Multiplication and Division in Polar Form

Cartesian arithmetic is dependable but blind: $(a + bi)(c + di)$(a+bi)(c+di) tells you the answer without telling you what happened. Polar form makes the geometry visible. The single most important fact in this lesson is that multiplying complex numbers multiplies their moduli and adds their arguments, while dividing divides the moduli and subtracts the arguments. From those two rules, powers, rotations and the whole machinery of De Moivre's theorem follow. Once you see complex multiplication as "scale and turn," a calculation that would take a page of bracket-expansion becomes a line of mental arithmetic — and the picture of what the answer means comes free.

---

Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, sitting between modulus–argument form (the previous lesson) and De Moivre's theorem (two lessons on). It is a blend of AO1 (apply the multiply-moduli / add-arguments rules accurately) and AO2 (interpret multiplication geometrically as a spiral similarity, and justify why the rules hold via the compound-angle formulae). It is also a quiet AO3 workhorse: rotation-by-multiplication is the cleanest way to handle "rotate this point/triangle about the origin" problems, which appear in transformation-geometry and vector contexts across the paper. Master the geometric picture here and De Moivre will feel like a restatement of something you already know.

---

Core theory: multiply moduli, add arguments

Write two complex numbers in modulus–argument (polar) form:

$z_1 = r_1(\cos\theta_1 + i\sin\theta_1), \qquad z_2 = r_2(\cos\theta_2 + i\sin\theta_2).$z1​=r1​(cosθ1​+isinθ1​),z2​=r2​(cosθ2​+isinθ2​).

The multiplication rule and its proof

The product is

$\boxed{\,z_1 z_2 = r_1 r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big).\,}$z1​z2​=r1​r2​(cos(θ1​+θ2​)+isin(θ1​+θ2​)).​

This is not an axiom — it is forced by the compound-angle formulae you met in A-Level Maths. Multiply out and group:

$\begin{aligned} z_1 z_2 &= r_1 r_2(\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2) \\ &= r_1 r_2\big[\cos\theta_1\cos\theta_2 + i\cos\theta_1\sin\theta_2 + i\sin\theta_1\cos\theta_2 + i^2\sin\theta_1\sin\theta_2\big] \\ &= r_1 r_2\big[(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)\big] \\ &= r_1 r_2\big[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big], \end{aligned}$z1​z2​​=r1​r2​(cosθ1​+isinθ1​)(cosθ2​+isinθ2​)=r1​r2​[cosθ1​cosθ2​+icosθ1​sinθ2​+isinθ1​cosθ2​+i2sinθ1​sinθ2​]=r1​r2​[(cosθ1​cosθ2​−sinθ1​sinθ2​)+i(sinθ1​cosθ2​+cosθ1​sinθ2​)]=r1​r2​[cos(θ1​+θ2​)+isin(θ1​+θ2​)],​

using $\cos(A+B) = \cos A\cos B - \sin A\sin B$cos(A+B)=cosAcosB−sinAsinB and $\sin(A+B) = \sin A\cos B + \cos A\sin B$sin(A+B)=sinAcosB+cosAsinB. The real part of the bracket is the cosine addition formula and the imaginary part is the sine addition formula — which is exactly why arguments add. In modulus–argument shorthand this is the clean pair

$|z_1 z_2| = |z_1|\,|z_2|, \qquad \arg(z_1 z_2) = \arg z_1 + \arg z_2 \;(\text{mod } 2\pi).$∣z1​z2​∣=∣z1​∣∣z2​∣,arg(z1​z2​)=argz1​+argz2​(mod 2π).

Geometric meaning: a spiral similarity

Multiplying every point $z$z by a fixed $w = r(\cos\theta + i\sin\theta)$w=r(cosθ+isinθ) does two things at once:

• it scales the distance from the origin by the factor $r = |w|$r=∣w∣;
• it rotates the point anticlockwise through the angle $\theta = \arg w$θ=argw.

A combined enlargement-and-rotation about a fixed point is called a spiral similarity (or rotational enlargement). The special case $|w| = 1$∣w∣=1 is a pure rotation: multiplying by $\cos\theta + i\sin\theta$cosθ+isinθ turns the plane by $\theta$θ without changing any distance. This is the mechanism behind "multiplication by $i$i is a quarter-turn," developed below. The other special case, $\arg w = 0$argw=0 (so $w$w is a positive real), is a pure enlargement by the factor $w$w. Every complex multiplication is one of these or a blend of the two — which is the single most useful mental picture in the whole strand.

The division rule

Dividing reverses multiplication, so moduli divide and arguments subtract:

$\boxed{\,\frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big),\,} \qquad (z_2 \neq 0).$z2​z1​​=r2​r1​​(cos(θ1​−θ2​)+isin(θ1​−θ2​)),​(z2​=0).

$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}, \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2 \;(\text{mod } 2\pi).$​z2​z1​​​=∣z2​∣∣z1​∣​,arg(z2​z1​​)=argz1​−argz2​(mod 2π).

One quick way to see this: the reciprocal of $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$z2​=r2​(cosθ2​+isinθ2​) is $z_2^{-1} = \tfrac{1}{r_2}(\cos(-\theta_2) + i\sin(-\theta_2))$z2−1​=r2​1​(cos(−θ2​)+isin(−θ2​)) — reciprocate the modulus, negate the argument — and then $z_1/z_2 = z_1 \cdot z_2^{-1}$z1​/z2​=z1​⋅z2−1​ applies the multiplication rule. (This also shows why the division rule is not a separate fact to memorise: it is the multiplication rule applied to $z_1$z1​ and $z_2^{-1}$z2−1​.) After any such calculation, bring the argument back into the principal range $(-\pi, \pi]$(−π,π] by adding or subtracting $2\pi$2π as needed; an answer of $\arg = \tfrac{7\pi}{6}$arg=67π​ should be reported as $-\tfrac{5\pi}{6}$−65π​. The polar approach is dramatically faster than Cartesian division (realising the denominator) whenever the moduli and arguments are known — compare one subtraction of angles against a full conjugate-multiplication and you can see why the exam rewards spotting when to convert.

Powers as repeated multiplication

Applying the multiplication rule with $z_1 = z_2 = z$z1​=z2​=z gives $z^2 = r^2(\cos 2\theta + i\sin 2\theta)$z2=r2(cos2θ+isin2θ); repeating,

$z^n = r^n\big(\cos n\theta + i\sin n\theta\big).$zn=rn(cosnθ+isinnθ).

This is De Moivre's theorem (proved formally two lessons on); for now it is just the multiplication rule used $n$n times, and it already lets us compute powers that would be brutal by binomial expansion.

Operation	Modulus	Argument
$z_1 z_2$z1​z2​	$r_1 r_2$r1​r2​ (multiply)	$\theta_1 + \theta_2$θ1​+θ2​ (add)
$z_1 / z_2$z1​/z2​	$r_1 / r_2$r1​/r2​ (divide)	$\theta_1 - \theta_2$θ1​−θ2​ (subtract)
$z^n$zn	$r^n$rn	$n\theta$nθ
$1/z$1/z	$1/r$1/r	$-\theta$−θ

---

Worked examples (with mark scheme)

Example 1 — multiplication

Given $z_1 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$z1​=2(cos3π​+isin3π​) and $z_2 = 3\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$z2​=3(cos6π​+isin6π​), find $z_1 z_2$z1​z2​ in the form $a + bi$a+bi.

$|z_1 z_2| = 2 \times 3 = 6, \qquad \arg(z_1 z_2) = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}. \quad (\text{M1 multiply moduli, add args})$∣z1​z2​∣=2×3=6,arg(z1​z2​)=3π​+6π​=2π​.(M1 multiply moduli, add args) $z_1 z_2 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6(0 + i\cdot 1) = 6i. \quad (\text{A1})$z1​z2​=6(cos2π​+isin2π​)=6(0+i⋅1)=6i.(A1)

(M1 for the two rules applied together; A1 for $6i$6i. Note how the answer drops out with no bracket-expansion at all.)

Example 2 — division

For the same $z_1, z_2$z1​,z2​, find $\dfrac{z_1}{z_2}$z2​z1​​ in the form $a + bi$a+bi.

$\left|\frac{z_1}{z_2}\right| = \frac{2}{3}, \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}. \quad (\text{M1 divide moduli, subtract args})$​z2​z1​​​=32​,arg(z2​z1​​)=3π​−6π​=6π​.(M1 divide moduli, subtract args) $\frac{z_1}{z_2} = \frac{2}{3}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \frac{2}{3}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{3}}{3} + \frac{1}{3}i. \quad (\text{A1})$z2​z1​​=32​(cos6π​+isin6π​)=32​(23​​+21​i)=33​​+31​i.(A1)

(M1 for the division rule; A1 for the exact Cartesian form. Equivalently $\tfrac{\sqrt 3}{3} = \tfrac{1}{\sqrt 3}$33​​=3​1​ — leave surds rationalised.)

Example 3 — a power via polar form

Find the modulus and argument of $(1 + i)^6$(1+i)6, and hence write it as $a + bi$a+bi.

First convert the base: $|1 + i| = \sqrt{1^2 + 1^2} = \sqrt 2$∣1+i∣=12+12​=2​ and $\arg(1 + i) = \frac{\pi}{4}$arg(1+i)=4π​ (first quadrant). Then

$(1 + i)^6 = (\sqrt 2)^6\left(\cos\frac{6\pi}{4} + i\sin\frac{6\pi}{4}\right). \quad (\text{M1 convert base; M1 apply } z^n = r^n\,\text{cis}\,n\theta)$(1+i)6=(2​)6(cos46π​+isin46π​).(M1 convert base; M1 apply zn=rncisnθ)

Now $(\sqrt 2)^6 = 2^3 = 8$(2​)6=23=8 and $\frac{6\pi}{4} = \frac{3\pi}{2}$46π​=23π​, so

$(1 + i)^6 = 8\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = 8(0 - i) = -8i. \quad (\text{A1})$(1+i)6=8(cos23π​+isin23π​)=8(0−i)=−8i.(A1)

So $|(1+i)^6| = 8$∣(1+i)6∣=8 and $\arg = \frac{3\pi}{2} \equiv -\frac{\pi}{2}$arg=23π​≡−2π​ in principal range. (Check by Cartesian: $(1+i)^2 = 2i$(1+i)2=2i, so $(1+i)^6 = (2i)^3 = 8i^3 = -8i$(1+i)6=(2i)3=8i3=−8i ✓.)

Example 4 — a larger power

Compute $(1 + \sqrt 3\, i)^3$(1+3​i)3.

$|1 + \sqrt 3\, i| = \sqrt{1 + 3} = 2, \qquad \arg = \arctan\frac{\sqrt 3}{1} = \frac{\pi}{3}. \quad (\text{M1 modulus and argument})$∣1+3​i∣=1+3​=2,arg=arctan13​​=3π​.(M1 modulus and argument) $(1 + \sqrt 3\, i)^3 = 2^3\big(\cos\pi + i\sin\pi\big) = 8(-1 + 0) = -8. \quad (\text{A1})$(1+3​i)3=23(cosπ+isinπ)=8(−1+0)=−8.(A1)

(M1 for the polar form of the base; A1 for $-8$−8. The answer is a negative real — a useful sanity flag, since $3\theta = \pi$3θ=π lands on the negative real axis.)

Example 5 — rotation of a figure

The triangle with vertices $z_1 = 1$z1​=1, $z_2 = 2 + i$z2​=2+i, $z_3 = i$z3​=i is rotated $90^\circ$90∘ anticlockwise about the origin. Find the image vertices.

A $90^\circ$90∘ anticlockwise rotation about $O$O is multiplication by $\cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = i$cos2π​+isin2π​=i. Multiply each vertex:

$z_1' = i(1) = i; \quad z_2' = i(2 + i) = 2i + i^2 = -1 + 2i; \quad z_3' = i(i) = i^2 = -1. \quad (\text{M1 multiply each by } i;\ \text{A1 all three})$z1′​=i(1)=i;z2′​=i(2+i)=2i+i2=−1+2i;z3′​=i(i)=i2=−1.(M1 multiply each by i; A1 all three)

So the image triangle has vertices $i$i, $-1 + 2i$−1+2i and $-1$−1. (M1 identifying the rotation as $\times i$×i; A1 for the three correct images. A rotation preserves side lengths — a quick check is that $|z_2 - z_1| = |z_2' - z_1'|$∣z2​−z1​∣=∣z2′​−z1′​∣.)

Example 6 — converting before multiplying

Find $(\sqrt 3 - i)(1 + i)$(3​−i)(1+i) by first converting each factor to polar form, and give the answer in the form $a + bi$a+bi.

Convert each factor. $\sqrt 3 - i$3​−i: modulus $\sqrt{3 + 1} = 2$3+1​=2, fourth-quadrant argument $-\frac{\pi}{6}$−6π​. $1 + i$1+i: modulus $\sqrt 2$2​, argument $\frac{\pi}{4}$4π​. (M1 both polar forms.) Multiply (moduli multiply, arguments add):

$(\sqrt 3 - i)(1 + i) = 2\sqrt 2\left(\cos\left(-\frac{\pi}{6} + \frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{6} + \frac{\pi}{4}\right)\right) = 2\sqrt 2\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right). \quad (\text{M1 multiply})$(3​−i)(1+i)=22​(cos(−6π​+4π​)+isin(−6π​+4π​))=22​(cos12π​+isin12π​).(M1 multiply)

Since $-\frac{\pi}{6} + \frac{\pi}{4} = \frac{-2\pi + 3\pi}{12} = \frac{\pi}{12}$−6π​+4π​=12−2π+3π​=12π​, the modulus is $2\sqrt 2$22​ and the argument $\frac{\pi}{12}$12π​. (A1.) As a Cartesian check, multiplying directly: $(\sqrt 3 - i)(1 + i) = \sqrt 3 + \sqrt 3\, i - i - i^2 = (\sqrt 3 + 1) + (\sqrt 3 - 1)i$(3​−i)(1+i)=3​+3​i−i−i2=(3​+1)+(3​−1)i — and indeed $2\sqrt 2\cos\frac{\pi}{12} = \sqrt 3 + 1$22​cos12π​=3​+1 and $2\sqrt 2\sin\frac{\pi}{12} = \sqrt 3 - 1$22​sin12π​=3​−1, giving the exact value of $\cos\frac{\pi}{12}$cos12π​ as a by-product. (M1 both polar forms; M1 multiply; A1 modulus–argument answer. This neat side-effect — polar multiplication producing exact values of $\cos\frac{\pi}{12}$cos12π​, $\sin\frac{\pi}{12}$sin12π​ — is a favourite exam twist.)

Example 7 — a quotient with a quadrant subtlety

Find $\dfrac{8\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)}{2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)}$2(cos65π​+isin65π​)8(cos43π​+isin43π​)​, giving the argument in $(-\pi, \pi]$(−π,π].

Moduli divide, arguments subtract:

$\text{modulus} = \frac{8}{2} = 4, \qquad \text{argument} = \frac{3\pi}{4} - \frac{5\pi}{6} = \frac{9\pi - 10\pi}{12} = -\frac{\pi}{12}. \quad (\text{M1 divide and subtract})$modulus=28​=4,argument=43π​−65π​=129π−10π​=−12π​.(M1 divide and subtract)

This already lies in $(-\pi, \pi]$(−π,π], so the quotient is $4\left(\cos\left(-\frac{\pi}{12}\right) + i\sin\left(-\frac{\pi}{12}\right)\right)$4(cos(−12π​)+isin(−12π​)). (M1 division rule; A1 with the principal argument. Had the subtraction strayed below $-\pi$−π, we would add $2\pi$2π; here no adjustment is needed.)

---

Specimen-style exam question

(specimen-style — not from any past paper)

The complex numbers $z_1$z1​ and $z_2$z2​ satisfy $|z_1| = 4$∣z1​∣=4, $\arg z_1 = \frac{2\pi}{3}$argz1​=32π​, $|z_2| = 2$∣z2​∣=2 and $\arg z_2 = -\frac{\pi}{4}$argz2​=−4π​. (a) Find $|z_1 z_2|$∣z1​z2​∣ and $\arg(z_1 z_2)$arg(z1​z2​). (b) Find $\left|\dfrac{z_1}{z_2}\right|$​z2​z1​​​ and $\arg\!\left(\dfrac{z_1}{z_2}\right)$arg(z2​z1​​), giving the argument in the range $(-\pi, \pi]$(−π,π].

(a) Moduli multiply, arguments add:

$|z_1 z_2| = 4 \times 2 = 8, \qquad \arg(z_1 z_2) = \frac{2\pi}{3} + \left(-\frac{\pi}{4}\right) = \frac{8\pi - 3\pi}{12} = \frac{5\pi}{12}.$∣z1​z2​∣=4×2=8,arg(z1​z2​)=32π​+(−4π​)=128π−3π​=125π​.

This already lies in $(-\pi, \pi]$(−π,π], so no adjustment is needed.

(b) Moduli divide, arguments subtract:

$\left|\frac{z_1}{z_2}\right| = \frac{4}{2} = 2, \qquad \arg\!\left(\frac{z_1}{z_2}\right) = \frac{2\pi}{3} - \left(-\frac{\pi}{4}\right) = \frac{8\pi + 3\pi}{12} = \frac{11\pi}{12}.$​z2​z1​​​=24​=2,arg(z2​z1​​)=32π​−(−4π​)=128π+3π​=1211π​.

Again $\frac{11\pi}{12} \le \pi$1211π​≤π, so it is already the principal argument. (Had the subtraction given, say, $\frac{13\pi}{12}$1213π​, we would subtract $2\pi$2π to report $-\frac{11\pi}{12}$−1211π​.)

---

Synoptic links