QED
Complex Numbers · step 5 of 13

Multiplying & dividing in polar form

In polar form multiplication becomes beautifully simple: moduli multiply and arguments add, so (r₁e^(iθ₁))(r₂e^(iθ₂)) = r₁r₂e^(i(θ₁+θ₂)). Division divides the moduli and subtracts the arguments. These are just the index laws, and they turn geometrically opaque Cartesian multiplication into a rotation-and-scaling statement.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

  1. Convert both numbers to polar formGet r and θ for each, with quadrant checks.
  2. Multiply: r₁r₂ and θ₁ + θ₂Moduli multiply, arguments add.
  3. Divide: r₁/r₂ and θ₁ − θ₂Moduli divide, arguments subtract.
  4. Bring the argument back into rangeIf the result leaves (−π, π], add or subtract 2π.

Worked example

With z = 2e^(iπ/3) and w = 3e^(iπ/6), find zw and z/w.

  1. Product modulus: 2 × 3 = 6. Product argument: π/3 + π/6 = π/2.
  2. So zw = 6e^(iπ/2) = 6i.
  3. Quotient modulus: 2/3. Quotient argument: π/3 − π/6 = π/6.
  4. So z/w = (2/3)e^(iπ/6) = (2/3)(√3/2 + i/2).

Answer. zw = 6i and z/w = (2/3)e^(iπ/6) = √3/3 + i/3.

Where marks get dropped

These are the specific errors that cost credit on multiplying & dividing in polar form questions — QED's rubric penalises each of them separately.

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Multiplying & dividing in polar form — frequently asked questions

Why do arguments add?

Because e^(iθ₁)·e^(iθ₂) = e^(i(θ₁+θ₂)) by the index law. Geometrically, multiplying by a complex number rotates by its argument.

What does multiplying by i do?

It rotates by π/2 anticlockwise, since i = e^(iπ/2) with modulus 1. Multiplying by i four times returns you to the start.

How does this lead to De Moivre?

Multiplying z by itself n times multiplies the modulus n times and adds the argument n times, giving (re^(iθ))ⁿ = rⁿe^(inθ).

The rest of Complex Numbers

Arithmetic, the Argand plane, polar form and De Moivre. Each subtopic below has its own method, worked example and mark-losing traps.

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