Polar and exponential form

Higher Level

z₁ has length 2, z₂ has length 3. Turn z₂ and watch the product: its arm is always the two angles added, and its length is always 6.

π/6

6|z₁z₂| = 2 × 3

1.571arg sum, rad

0 + 6iin a + bi

Two numbers in, one rotation and one stretch out.

The three forms

z = a + bi (Cartesian) = r(cosθ + i sinθ) = r cisθ (polar) = re^iθ (exponential, or Euler form). Same number, three descriptions. r = |z| and θ = arg z.

2 cis(π/3)= 2cos(π/3) + 2i sin(π/3) = 1 + 1.732i multiply→ r₁r₂ cis(θ₁ + θ₂) 2cis(π/3) × 3cis(π/6)= 6 cis(π/2) = 6i divide→ r₁r₂ cis(θ₁ − θ₂)

In Cartesian form that product takes four terms and an i². In polar it takes one line. That is the entire reason this form exists, and it is why anything involving repeated rotation, from alternating current to computer graphics, is written this way.

Euler's identity

Put r = 1 and θ = π into re^iθ and you get e^iπ = −1, usually written e^iπ + 1 = 0. In the picture it is unremarkable: a half turn from 1 lands on −1. Written down it ties together e, i, π, 1 and 0 in one line.

Worked example

z = 1 + i√3.

Write z in polar form and use it to find z⁶.

|z| = √(1 + 3) = 2Pythagoras on the real and imaginary parts.
arg z = arctan(√3 / 1) = π/3Check the quadrant: both parts are positive, so it is in the first and arctan can be trusted here.
z = 2 cis(π/3)Modulus and argument, nothing else needed.
z⁶ = 2⁶ cis(6 × π/3) = 64 cis(2π)De Moivre: the modulus takes the power, the argument multiplies.
= 64, a real number2π is a full turn back to the positive real axis. Doing the same in Cartesian form means expanding a bracket to the sixth power, which is why this notation exists.

Your turn

1. z₁ = 2cis(π/3) and z₂ = 3cis(π/6). Give |z₁z₂|.

2. For the same two, give arg(z₁z₂) in radians to 3 decimal places.

3. Give the real part of 2cis(π/3).

Where the marks go

Working in radians. A calculator in degrees gives wrong moduli and arguments silently, all the way through.

Multiplying moduli and adding arguments, rather than converting to Cartesian and expanding, when the question is set in polar form.

Giving the argument in the range the question asks for, usually −π < θ ≤ π, which may mean adding or subtracting 2π at the end.

Come back to this

These three are deliberately not all about this page. In an examination the hardest step is deciding which method applies, and questions that arrive straight after the method never make you decide. Answer from memory before you reveal anything.

Come back to these in a week, and again in a month.

Which method?

Find the real part of (3 + 4i)(1 + 2i).

Cartesian multiplication: expand and use i² = −1, giving −5. In polar form you would instead multiply moduli and add arguments.
Which method?

200000 borrowed at 6%, repaid monthly over 25 years. At payment 150 of 300, roughly how much is still owed?

Amortization, and not half. About 135754. The balance does not fall in a straight line.
Which method?

Solve 3 × 2ᵗ = 96.

Isolate the power first: 2ᵗ = 32, so t = 5. Taking logs before dividing by 3 turns one line into three.

The same idea elsewhere

1.12, complex numbers: The modulus and argument were defined there; this page uses them as coordinates.
5.13, kinematics: Rotating quantities are how circular motion and waves get described.

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