Skip to main content

Multiplication and division in polar form (WACE Year 12 Mathematics Specialist Unit 3)

Syllabus dot point

“Multiply and divide complex numbers in polar form, interpreting the result as rotation and scaling”

7 min read

Quick answer

In polar form multiplication multiplies moduli and adds arguments, while division divides moduli and subtracts arguments: z1z2=r1r2 cis(θ1+θ2)z_1 z_2 = r_1 r_2\,\text{cis}(\theta_1 + \theta_2) and z1z2=r1r2 cis(θ1−θ2)\tfrac{z_1}{z_2} = \tfrac{r_1}{r_2}\,\text{cis}(\theta_1 - \theta_2). So multiplying by a complex number rotates by its argument and scales by its modulus.

Jump to a section
  1. What this dot point is asking
  2. The multiplication rule
  3. The division rule
  4. Geometric interpretation
  5. Deriving the rule from the angle-sum identities
  6. Transformations of the plane
  7. Why polar form helps
  8. Exam-style questions

What this dot point is asking

SCSA wants you to combine complex numbers in modulus-argument form and read off the geometric meaning: multiplication as a rotation plus dilation about the origin.

The multiplication rule

Write z1=r1 cis θ1z_1 = r_1\,\text{cis}\,\theta_1 and z2=r2 cis θ2z_2 = r_2\,\text{cis}\,\theta_2. Expanding the product and using the angle-sum identities cos⁡(θ1+θ2)\cos(\theta_1 + \theta_2) and sin⁡(θ1+θ2)\sin(\theta_1 + \theta_2) gives

Product in polar form

z1z2=r1r2 cis(θ1+θ2).z_1 z_2 = r_1 r_2\,\text{cis}(\theta_1 + \theta_2).

So ∣z1z2∣=∣z1∣ ∣z2∣|z_1 z_2| = |z_1|\,|z_2| and arg⁡(z1z2)=arg⁡z1+arg⁡z2\arg(z_1 z_2) = \arg z_1 + \arg z_2 (up to a multiple of 2π2\pi, since the result may need adjusting back into the principal range).

The division rule

Dividing reverses the process:

z1z2=r1r2 cis(θ1−θ2).\frac{z_1}{z_2} = \frac{r_1}{r_2}\,\text{cis}(\theta_1 - \theta_2).

Moduli divide and arguments subtract. As a special case 1z=1r cis(−θ)\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta), the reciprocal having reciprocal modulus and negated argument.

Geometric interpretation

Multiplication as rotation and scaling

Multiplying zz by r cis αr\,\text{cis}\,\alpha rotates the point zz anticlockwise by angle α\alpha about the origin and scales its distance from the origin by the factor rr.

Two clean cases worth memorising: multiplying by i=cis π2i = \text{cis}\,\tfrac{\pi}{2} rotates by a quarter turn anticlockwise with no scaling, and multiplying by −1=cis π-1 = \text{cis}\,\pi rotates by a half turn. The conjugate in polar form is r cis θ‾=r cis(−θ)\overline{r\,\text{cis}\,\theta} = r\,\text{cis}(-\theta), a reflection in the real axis.

Deriving the rule from the angle-sum identities

The product rule is not a definition to memorise but a consequence of the trigonometric angle-sum identities. Multiplying r1(cos⁡θ1+isin⁡θ1)r_1(\cos\theta_1 + i\sin\theta_1) by r2(cos⁡θ2+isin⁡θ2)r_2(\cos\theta_2 + i\sin\theta_2) and expanding gives a real part r1r2(cos⁡θ1cos⁡θ2−sin⁡θ1sin⁡θ2)r_1 r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) and an imaginary part r1r2(sin⁡θ1cos⁡θ2+cos⁡θ1sin⁡θ2)r_1 r_2(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2). The bracketed expressions are exactly cos⁡(θ1+θ2)\cos(\theta_1 + \theta_2) and sin⁡(θ1+θ2)\sin(\theta_1 + \theta_2), so the product is r1r2 cis(θ1+θ2)r_1 r_2\,\text{cis}(\theta_1 + \theta_2). Knowing this derivation is worth marks when a question asks you to justify the rule rather than just apply it.

Transformations of the plane

Because multiplication by r cis αr\,\text{cis}\,\alpha rotates and scales, complex multiplication is a compact way to describe geometric transformations about the origin. A rotation by α\alpha alone is multiplication by cis α\text{cis}\,\alpha (modulus 11); a dilation by factor kk alone is multiplication by the real number kk; and a spiral similarity combining both is multiplication by k cis αk\,\text{cis}\,\alpha. This viewpoint connects the complex-numbers topic to the geometry of the Argand plane and to the later study of curves and regions.

Why polar form helps

Cartesian multiplication is algebraically heavy, but polar form reduces it to one product and one sum. This is also the engine behind de Moivre's theorem, which is just repeated multiplication of equal factors.

Worked example

Compute z1z2z_1 z_2 and z1z2\dfrac{z_1}{z_2} for z1=4 cis2π3z_1 = 4\,\text{cis}\tfrac{2\pi}{3}, z2=2 cisπ6z_2 = 2\,\text{cis}\tfrac{\pi}{6}

Step 1: Compute the product by multiplying moduli and adding arguments.

In polar form, multiplication combines the two moduli by multiplication and the two arguments by addition. This is far quicker than expanding in Cartesian form and reflects the geometric interpretation of multiplication as a rotation and scaling:

z1z2=(4)(2) cis ⁣(2π3+π6)=8 cis ⁣(5π6).z_1 z_2 = (4)(2)\,\text{cis}\!\left(\frac{2\pi}{3} + \frac{\pi}{6}\right) = 8\,\text{cis}\!\left(\frac{5\pi}{6}\right).

Step 2: Compute the quotient by dividing moduli and subtracting arguments.

Division is the reverse operation: moduli divide and arguments subtract. Because the result is on the unit circle (modulus 22) at argument π2\tfrac{\pi}{2}, it can be converted directly to Cartesian form:

z1z2=42 cis ⁣(2π3−π6)=2 cis ⁣(π2)=2i.\frac{z_1}{z_2} = \frac{4}{2}\,\text{cis}\!\left(\frac{2\pi}{3} - \frac{\pi}{6}\right) = 2\,\text{cis}\!\left(\frac{\pi}{2}\right) = 2i.

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

Common traps
Adding moduli or multiplying arguments
It is the other way round: moduli multiply, arguments add. This reversal is the most frequent error.
Leaving the argument out of range
If a sum like 2π3+3π4\tfrac{2\pi}{3} + \tfrac{3\pi}{4} exceeds π\pi, subtract 2π2\pi to return to the principal value.
Forgetting to scale
Multiplying by r cis αr\,\text{cis}\,\alpha with r≠1r \neq 1 changes the distance from the origin, not just the angle.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2023-style5 marks
Calculator-free. Let z=1+iz = 1 + i. (a) Write zz in polar form. (b) Hence find z×z‾z \times \overline{z} and zz‾\dfrac{z}{\overline{z}} in polar form, and interpret zz‾\dfrac{z}{\overline{z}} geometrically.
Show worked answer →

Combines conversion, conjugate and the polar rules.

(a) ∣z∣=1+1=2|z| = \sqrt{1 + 1} = \sqrt{2}; the point (1,1)(1, 1) is in the first quadrant so arg⁡z=π4\arg z = \dfrac{\pi}{4}. Thus z=2 cis π4z = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}.

(b) The conjugate is z‾=2 cis ⁣(−π4)\overline{z} = \sqrt{2}\,\text{cis}\!\left(-\dfrac{\pi}{4}\right). Product: multiply moduli, add arguments: zz‾=2 cis 0=2z\overline{z} = 2\,\text{cis}\,0 = 2, which is ∣z∣2|z|^2 as expected. Quotient: divide moduli, subtract arguments: zz‾=1⋅cis ⁣(π4−(−π4))=cis π2=i\dfrac{z}{\overline{z}} = 1 \cdot \text{cis}\!\left(\dfrac{\pi}{4} - \left(-\dfrac{\pi}{4}\right)\right) = \text{cis}\,\dfrac{\pi}{2} = i.

Geometrically zz‾\dfrac{z}{\overline{z}} has modulus 11 and rotates by π2\dfrac{\pi}{2}, a pure quarter-turn anticlockwise about the origin. Markers reward the polar form, both results, and the rotation interpretation.

WACE 2020-style4 marks
Calculator-assumed. The point PP represents z=3 cis π6z = 3\,\text{cis}\,\dfrac{\pi}{6}. Find the complex number represented by the image of PP after a rotation of π3\dfrac{\pi}{3} anticlockwise about the origin followed by a dilation of factor 22.
Show worked answer →

A transformation expressed as a polar multiplication.

A rotation of π3\dfrac{\pi}{3} and dilation of factor 22 about the origin is exactly multiplication by 2 cis π32\,\text{cis}\,\dfrac{\pi}{3}.

So the image is z′=3 cis π6×2 cis π3=6 cis ⁣(π6+π3)=6 cis π2=6iz' = 3\,\text{cis}\,\dfrac{\pi}{6} \times 2\,\text{cis}\,\dfrac{\pi}{3} = 6\,\text{cis}\!\left(\dfrac{\pi}{6} + \dfrac{\pi}{3}\right) = 6\,\text{cis}\,\dfrac{\pi}{2} = 6i.

Markers reward identifying the transformation as multiplication by 2 cis π32\,\text{cis}\,\dfrac{\pi}{3}, adding the arguments, and the final value 6i6i.

Practise this

Sources & how we know this

ExamExplained