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Why does multiplying complex numbers rotate and scale, and how does polar form make this obvious?

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

WACE Specialist Unit 3 polar form arithmetic: multiplying moduli and adding arguments, dividing moduli and subtracting arguments, the geometric interpretation as rotation and dilation, and the conjugate in polar form, with a worked example.

Reviewed by: AI editorial process; not yet individually human-reviewed

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  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

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=r1cisθ1z_1 = r_1\,\text{cis}\,\theta_1 and z2=r2cisθ2z_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

So z1z2=z1z2|z_1 z_2| = |z_1|\,|z_2| and arg(z1z2)=argz1+argz2\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=r1r2cis(θ1θ2).\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=1rcis(θ)\tfrac{1}{z} = \tfrac{1}{r}\,\text{cis}(-\theta), the reciprocal having reciprocal modulus and negated argument.

Geometric interpretation

Two clean cases worth memorising: multiplying by i=cisπ2i = \text{cis}\,\tfrac{\pi}{2} rotates by a quarter turn anticlockwise with no scaling, and multiplying by 1=cisπ-1 = \text{cis}\,\pi rotates by a half turn. The conjugate in polar form is rcisθ=rcis(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(cosθ1+isinθ1)r_1(\cos\theta_1 + i\sin\theta_1) by r2(cosθ2+isinθ2)r_2(\cos\theta_2 + i\sin\theta_2) and expanding gives a real part r1r2(cosθ1cosθ2sinθ1sinθ2)r_1 r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) and an imaginary part r1r2(sinθ1cosθ2+cosθ1sinθ2)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 r1r2cis(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 rcisαr\,\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 cisα\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 kcisαk\,\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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20235 marksCalculator-free. Let z=1+iz = 1 + i. (a) Write zz in polar form. (b) Hence find z×zz \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 argz=π4\arg z = \dfrac{\pi}{4}. Thus z=2cisπ4z = \sqrt{2}\,\text{cis}\,\dfrac{\pi}{4}.

(b) The conjugate is z=2cis ⁣(π4)\overline{z} = \sqrt{2}\,\text{cis}\!\left(-\dfrac{\pi}{4}\right). Product: multiply moduli, add arguments: zz=2cis0=2z\overline{z} = 2\,\text{cis}\,0 = 2, which is z2|z|^2 as expected. Quotient: divide moduli, subtract arguments: zz=1cis ⁣(π4(π4))=cisπ2=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 20204 marksCalculator-assumed. The point PP represents z=3cisπ6z = 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 2cisπ32\,\text{cis}\,\dfrac{\pi}{3}.

So the image is z=3cisπ6×2cisπ3=6cis ⁣(π6+π3)=6cisπ2=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 2cisπ32\,\text{cis}\,\dfrac{\pi}{3}, adding the arguments, and the final value 6i6i.

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