Modulus and Argument (Polar Form)

Cartesian form $a + bi$a+bi is perfect for adding complex numbers, but clumsy for multiplying, dividing and taking powers. Polar (modulus–argument) form fixes this by describing $z$z with two numbers — how far it is from the origin (the modulus $r = |z|$r=∣z∣) and which direction it points (the argument $\theta = \arg z$θ=argz). In this language multiplication becomes "multiply the lengths, add the angles," which is the gateway to De Moivre's theorem, $n$nth roots and the exponential form $re^{i\theta}$reiθ. Think of it as switching from "rectangular coordinates" (real and imaginary parts) to "polar coordinates" (radius and angle) for the same point on the Argand diagram — exactly the change of viewpoint you met for plane curves in A-Level Maths, now applied to complex numbers.

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Where this sits in AQA 7367

This is compulsory pure content for Papers 1 & 2, and the pivot of the whole strand. It is strongly AO1 (compute $|z|$∣z∣ and $\arg z$argz, convert both ways) with crucial AO2 (choose the correct quadrant, work within the principal range, justify the modulus/argument rules). Mastery here is the prerequisite for multiplication/division in polar form, De Moivre's theorem, roots of unity and the exponential form — every one of those topics is stated in terms of $r$r and $\theta$θ.

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Core theory: length and direction

The modulus

The modulus of $z = a + bi$z=a+bi is its distance from the origin on the Argand diagram — the length of the position vector to the point $(a, b)$(a,b). By Pythagoras,

$|z| = \sqrt{a^2 + b^2}.$∣z∣=a2+b2​.

It generalises the absolute value of a real number: for real $z = a$z=a, $|z| = \sqrt{a^2} = |a|$∣z∣=a2​=∣a∣, the usual "distance from $0$0." The modulus is always a non-negative real number, even though $z$z itself is complex. Because it measures distance, the modulus is the quantity you use whenever a question asks "how far," "how big," or "the length of" — and, crucially, it is the only sense in which one complex number can be meaningfully compared in size with another (via their moduli).

| $z$z | $|z|$∣z∣ | |---|---| | $3 + 4i$3+4i | $\sqrt{9 + 16} = 5$9+16​=5 | | $-5 + 12i$−5+12i | $\sqrt{25 + 144} = 13$25+144​=13 | | $1 - i$1−i | $\sqrt{2}$2​ | | $7$7 | $7$7 | | $-3i$−3i | $3$3 |

It satisfies the key algebraic properties (all provable from $|z|^2 = z\bar z$∣z∣2=zzˉ):

Property	Formula
Non-negativity	$
Conjugate	$
Modulus squared	$
Product	$
Quotient	$\left
Triangle inequality	$

The product rule has a clean one-line proof using $|z|^2 = z\bar z$∣z∣2=zzˉ and $\overline{z_1 z_2} = \bar z_1\bar z_2$z1​z2​​=zˉ1​zˉ2​:

$|z_1 z_2|^2 = (z_1 z_2)\overline{(z_1 z_2)} = z_1 z_2\,\bar z_1\bar z_2 = (z_1\bar z_1)(z_2\bar z_2) = |z_1|^2|z_2|^2,$∣z1​z2​∣2=(z1​z2​)(z1​z2​)​=z1​z2​zˉ1​zˉ2​=(z1​zˉ1​)(z2​zˉ2​)=∣z1​∣2∣z2​∣2,

and taking (positive) square roots gives $|z_1 z_2| = |z_1||z_2|$∣z1​z2​∣=∣z1​∣∣z2​∣. The quotient rule follows by applying this to $z_1 = \tfrac{z_1}{z_2}\cdot z_2$z1​=z2​z1​​⋅z2​. These multiplicative properties are what make polar form so efficient for products and powers.

The argument

For $z \neq 0$z=0, the argument $\arg z = \theta$argz=θ is the angle the position vector makes with the positive real axis, measured anticlockwise. It is defined only up to multiples of $2\pi$2π; the principal argument is the unique value in

$-\pi < \theta \leq \pi.$−π<θ≤π.

To find it, compute the acute reference angle $\alpha = \arctan\dfrac{|b|}{|a|}$α=arctan∣a∣∣b∣​ and then place it in the correct quadrant — never just type $\arctan(b/a)$arctan(b/a) into a calculator, which only ever returns answers in $(-\tfrac{\pi}{2}, \tfrac{\pi}{2})$(−2π​,2π​) and so gets quadrants 2 and 3 wrong.

Quadrant of $(a,b)$(a,b)	$\arg z$argz
1st ($a>0, b>0$a>0,b>0)	$\alpha$α
2nd ($a<0, b>0$a<0,b>0)	$\pi - \alpha$π−α
3rd ($a<0, b<0$a<0,b<0)	$\alpha - \pi$α−π
4th ($a>0, b<0$a>0,b<0)	$-\alpha$−α

Axis cases: $\arg(r) = 0$arg(r)=0, $\arg(-r) = \pi$arg(−r)=π (for $r>0$r>0); $\arg(ki) = \tfrac{\pi}{2}$arg(ki)=2π​, $\arg(-ki) = -\tfrac{\pi}{2}$arg(−ki)=−2π​ (for $k>0$k>0). The argument of $0$0 is undefined — the zero vector has no direction.

The exact-angle values you will need most often come from the standard triangles; it is worth having them at your fingertips:

Reference angle $\alpha$α	$\tan\alpha$tanα	typical $a + bi$a+bi
$\tfrac{\pi}{6}$6π​	$\tfrac{1}{\sqrt 3}$3​1​	$\sqrt 3 + i$3​+i
$\tfrac{\pi}{4}$4π​	$1$1	$1 + i$1+i
$\tfrac{\pi}{3}$3π​	$\sqrt 3$3​	$1 + \sqrt 3\,i$1+3​i

Recognising, say, that $b/a = \sqrt 3$b/a=3​ instantly gives a reference angle of $\tfrac{\pi}{3}$3π​ saves time and avoids calculator-rounding. For non-standard ratios you will of course use $\arctan$arctan and quote a decimal (in radians) to the required accuracy.

The diagram shows $z = 1 + \sqrt 3\,i$z=1+3​i: modulus $r = 2$r=2 (the length of the position vector) and argument $\theta = \tfrac{\pi}{3}$θ=3π​ (the angle from the positive real axis), with the right triangle of sides $a = 1$a=1, $b = \sqrt 3$b=3​.

Modulus–argument (polar) form

Reading the right triangle gives $a = r\cos\theta$a=rcosθ and $b = r\sin\theta$b=rsinθ, so every non-zero $z$z can be written

$z = r(\cos\theta + i\sin\theta), \qquad r = |z|,\ \theta = \arg z.$z=r(cosθ+isinθ),r=∣z∣, θ=argz.

Cartesian $\to$→ polar: find $r = \sqrt{a^2 + b^2}$r=a2+b2​, then $\theta$θ by the quadrant method. Polar $\to$→ Cartesian: evaluate $a = r\cos\theta$a=rcosθ, $b = r\sin\theta$b=rsinθ.

There is an even more compact notation, exponential (Euler) form, which you will study fully alongside De Moivre's theorem:

$z = re^{i\theta}, \qquad \text{where } e^{i\theta} = \cos\theta + i\sin\theta.$z=reiθ,where eiθ=cosθ+isinθ.

In this form the multiplicative rules become the ordinary index laws: $r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2} = r_1 r_2\,e^{i(\theta_1 + \theta_2)}$r1​eiθ1​⋅r2​eiθ2​=r1​r2​ei(θ1​+θ2​) and $(re^{i\theta})^n = r^n e^{in\theta}$(reiθ)n=rneinθ. It is the same modulus–argument information written more economically — worth meeting now so that "multiply moduli, add arguments" feels inevitable rather than arbitrary.

The argument obeys multiplicative rules that make polar form so powerful:

Property	Formula
Product	$\arg(z_1 z_2) = \arg z_1 + \arg z_2$arg(z1​z2​)=argz1​+argz2​
Quotient	$\arg\!\left(\dfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$arg(z2​z1​​)=argz1​−argz2​
Conjugate	$\arg(\bar z) = -\arg z$arg(zˉ)=−argz
Negative	$\arg(-z) = \arg z \pm \pi$arg(−z)=argz±π

Combined with $|z_1 z_2| = |z_1||z_2|$∣z1​z2​∣=∣z1​∣∣z2​∣: multiply moduli, add arguments (and divide moduli, subtract arguments) — adjusting by $\pm 2\pi$±2π if needed to land in the principal range.

The "add arguments" rule is not magic — it is the angle-addition formulae in disguise. Writing $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$z1​=r1​(cosθ1​+isinθ1​) and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$z2​=r2​(cosθ2​+isinθ2​) and multiplying out,

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

using $\cos(A+B)$cos(A+B) and $\sin(A+B)$sin(A+B). So the product has modulus $r_1 r_2$r1​r2​ and argument $\theta_1 + \theta_2$θ1​+θ2​ — moduli multiply, arguments add. (This is the seed of De Moivre's theorem, where repeating the rule $n$n times gives $z^n = r^n(\cos n\theta + i\sin n\theta)$zn=rn(cosnθ+isinnθ).)

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Worked examples (with mark scheme)

Example 1 — modulus and argument (first quadrant)

Find $|z|$∣z∣ and $\arg z$argz for $z = 1 + \sqrt 3\,i$z=1+3​i.

$|z| = \sqrt{1 + 3} = 2. \quad (\text{M1 A1})$∣z∣=1+3​=2.(M1 A1) $a, b > 0 \Rightarrow \text{1st quadrant}; \;\; \alpha = \arctan\frac{\sqrt 3}{1} = \frac{\pi}{3} = \arg z. \quad (\text{M1 quadrant, A1})$a,b>0⇒1st quadrant;α=arctan13​​=3π​=argz.(M1 quadrant, A1)

(M1/A1 modulus; M1 for identifying the quadrant; A1 for $\tfrac{\pi}{3}$3π​.)

Example 2 — third-quadrant argument (sign care)

Find $|z|$∣z∣ and $\arg z$argz for $z = -3 - 3i$z=−3−3i.

$|z| = \sqrt{9 + 9} = 3\sqrt 2. \quad (\text{A1})$∣z∣=9+9​=32​.(A1) $a, b < 0 \Rightarrow \text{3rd quadrant}; \;\; \alpha = \arctan\frac{3}{3} = \frac{\pi}{4}; \;\; \arg z = \alpha - \pi = -\frac{3\pi}{4}. \quad (\text{M1 A1})$a,b<0⇒3rd quadrant;α=arctan33​=4π​;argz=α−π=−43π​.(M1 A1)

(M1 for the quadrant adjustment; A1 for $-\tfrac{3\pi}{4}$−43π​ — note it is negative, in $(-\pi, 0)$(−π,0), not $\tfrac{5\pi}{4}$45π​.)

Example 3 — convert to polar form

Express $z = -1 + i$z=−1+i in modulus–argument form.

$r = \sqrt{1 + 1} = \sqrt 2; \quad a<0, b>0 \Rightarrow \text{2nd quadrant}, \; \alpha = \frac{\pi}{4}, \; \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.$r=1+1​=2​;a<0,b>0⇒2nd quadrant,α=4π​,θ=π−4π​=43π​. $z = \sqrt 2\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right). \quad (\text{M1 } r;\ \text{M1 } \theta;\ \text{A1 form})$z=2​(cos43π​+isin43π​).(M1 r; M1 θ; A1 form)

Example 4 — fourth-quadrant argument

Find $|z|$∣z∣ and $\arg z$argz for $z = \sqrt 3 - i$z=3​−i.

$|z| = \sqrt{3 + 1} = 2. \quad (\text{A1})$∣z∣=3+1​=2.(A1) $a > 0, b < 0 \Rightarrow \text{4th quadrant};\;\; \alpha = \arctan\frac{1}{\sqrt 3} = \frac{\pi}{6};\;\; \arg z = -\alpha = -\frac{\pi}{6}. \quad (\text{M1 A1})$a>0,b<0⇒4th quadrant;α=arctan3​1​=6π​;argz=−α=−6π​.(M1 A1)

(M1 for the fourth-quadrant adjustment; A1 for $-\tfrac{\pi}{6}$−6π​. In the fourth quadrant the principal argument is simply $-\alpha$−α — negative, but already in $(-\pi, \pi]$(−π,π].)

Example 5 — polar to Cartesian with exact surds

Write $z = 4\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$z=4(cos32π​+isin32π​) in the form $a + bi$a+bi.

$a = 4\cos\frac{2\pi}{3} = 4\cdot\left(-\frac12\right) = -2, \qquad b = 4\sin\frac{2\pi}{3} = 4\cdot\frac{\sqrt 3}{2} = 2\sqrt 3.$a=4cos32π​=4⋅(−21​)=−2,b=4sin32π​=4⋅23​​=23​. $\therefore\; z = -2 + 2\sqrt 3\,i. \quad (\text{M1 exact values; A1})$∴z=−2+23​i.(M1 exact values; A1)

(M1 for the exact values of $\cos\frac{2\pi}{3}$cos32π​ and $\sin\frac{2\pi}{3}$sin32π​; A1 for the Cartesian form. Check: $|z| = \sqrt{4 + 12} = 4$∣z∣=4+12​=4 ✓, recovering the modulus.)

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Specimen-style exam question

(specimen-style — not from any past paper)

Given $z = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$z=2(cos6π​+isin6π​) and $w = 3\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$w=3(cos4π​+isin4π​), express $zw$zw and $\dfrac{z}{w}$wz​ in modulus–argument form.

Multiply moduli, add arguments:

$|zw| = 6, \quad \arg(zw) = \frac{\pi}{6} + \frac{\pi}{4} = \frac{2\pi + 3\pi}{12} = \frac{5\pi}{12},$∣zw∣=6,arg(zw)=6π​+4π​=122π+3π​=125π​, $zw = 6\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right).$zw=6(cos125π​+isin125π​).

Divide moduli, subtract arguments:

$\left|\frac{z}{w}\right| = \frac{2}{3}, \quad \arg\frac{z}{w} = \frac{\pi}{6} - \frac{\pi}{4} = -\frac{\pi}{12},$​wz​​=32​,argwz​=6π​−4π​=−12π​, $\frac{z}{w} = \frac{2}{3}\left(\cos\!\left(-\frac{\pi}{12}\right) + i\sin\!\left(-\frac{\pi}{12}\right)\right).$wz​=32​(cos(−12π​)+isin(−12π​)).

Both arguments already lie in $(-\pi, \pi]$(−π,π], so no adjustment is needed.

(continued) Hence write $z^2$z2 in modulus–argument form.

Squaring multiplies the modulus by itself and doubles the argument:

$|z^2| = 2^2 = 4, \qquad \arg(z^2) = 2\times\frac{\pi}{6} = \frac{\pi}{3},$∣z2∣=22=4,arg(z2)=2×6π​=3π​, $z^2 = 4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right).$z2=4(cos3π​+isin3π​).