NCEA Level 3Polar formAS 91577

Polar form by hand: modulus, argument and quadrants

What this lesson covers

  • Modulus is a distance, so it is never negative; argument is an angle measured from the positive real axis.
  • The quadrant decides the argument, so a sketch comes before the calculator.
Key idea

Polar form: a distance and a direction

  • x + iy is the point (x, y).
  • Modulus r: the distance from the origin.
  • Argument θ: the angle from the real axis.
Worked example

The Argand diagram

  • z = 1 + √3i: first quadrant.
  • w = −2 + 2i: second quadrant.
  • u = 1 − i: fourth quadrant.
The Argand diagram
Worked example

z = 1 + √3i

  • Sketch: name the quadrant.
  • Modulus, then argument.
  • Check by converting back.
  1. z = 1 + √3i: x > 0, y > 0first quadrant
  2. r = √(1² + (√3)²)modulus
  3. = √(1 + 3) = √4 = 2
  4. θ = tan⁻¹(√3/1) = π/3argument
  5. first quadrant: keep π/3
  6. z = 2 cis(π/3)
  7. 2(1/2) + 2(√3/2)i = 1 + √3i
Worked example

w = −2 + 2i

  • Sketch: second quadrant.
  • Calculator says −π/4.
  • Then π − reference angle.
  1. w = −2 + 2i: x < 0, y > 0second quadrant
  2. r = √((−2)² + 2²)modulus
  3. = √(4 + 4) = √8 = 2√2
  4. calculator: tan⁻¹(−1) = −π/4fourth quadrant
  5. reference: tan⁻¹(2/2) = π/4
  6. θ = π − π/4 = 3π/4second quadrant
  7. w = 2√2 cis(3π/4)
  8. 2√2(−√2/2) = −2, 2√2(√2/2) = 2
Worked example

The 2025 error: the lost minus sign

  • Below the axis: θ < 0.
  • Measured clockwise.
  • Range: −π < θ ≤ π.
  1. u = 1 − i: x > 0, y < 0fourth quadrant
  2. r = √(1² + (−1)²) = √2modulus
  3. reference: tan⁻¹(1/1) = π/4
  4. θ = π/4sign lost
  5. θ = −π/4below the real axis
  6. u = √2 cis(−π/4)
  7. −π < θ ≤ πthe usual range
Worked example

The modulus, written right

  • Square b, not bi.
  • A modulus: never negative.
  1. |3 + 4i| = √(3² + 4²)no i
  2. = √(9 + 16) = √25 = 5
  3. √(3² + (4i)²) = √(9 − 16)squared the i
  4. = √(−7)not a distance
  5. |z| ≥ 0never negative
Key idea

Exact values, no calculator

  • Converting by hand needs exact values.
  • π/6, π/4 and π/3 cover most questions.
θcos θsin θ
π/6√3/21/2
π/4√2/2√2/2
π/31/2√3/2
Worked example

4 cis(−2π/3) in rectangular form

  • x = r cos θ, y = r sin θ.
  • Exact values only.
  1. x = r cos θ, y = r sin θ
  2. −2π/3: third quadrantreference π/3
  3. cos(−2π/3) = −1/2
  4. sin(−2π/3) = −√3/2
  5. x = 4(−1/2) = −2
  6. y = 4(−√3/2) = −2√3
  7. 4 cis(−2π/3) = −2 − 2√3i
  8. √(4 + 12) = √16 = 4modulus
Worked example

Multiplying in polar form

  • Multiply the moduli.
  • Add the arguments.
  • Bring it into range.
  1. z = 2 cis(π/3), v = 3 cis(5π/6)
  2. |zv| = 2 × 3 = 6multiply moduli
  3. arg = π/3 + 5π/6add arguments
  4. = 2π/6 + 5π/6 = 7π/6
  5. 7π/6 > πout of range
  6. 7π/6 − 12π/6 = −5π/6subtract 2π
  7. zv = 6 cis(−5π/6)
Worked example

Dividing, and the answer form

  • Divide the moduli.
  • Subtract the arguments.
  • Then give the form asked.
  1. |z/v| = 2 ÷ 3 = 2/3divide moduli
  2. arg = π/3 − 5π/6subtract arguments
  3. = 2π/6 − 5π/6 = −3π/6
  4. = −π/2in range
  5. z/v = (2/3) cis(−π/2)
  6. cos(−π/2) = 0, sin(−π/2) = −1
  7. (2/3)(0 − i) = −(2/3)ithe form asked
Key idea

Sketch before the calculator

  • Sketch the point, name the quadrant.
  • Reference angle from the sizes of x and y.
  • Adjust for the quadrant, check the sign.
Key idea

Where the marks are

  • Achieved: modulus, argument, converting.
  • Merit: use that a modulus is positive.
  • Excellence: quotients, argument and parts.
Summary

What to take away

  • Sketch first: the quadrant decides the argument.
Sketch firstwhich quadrant?
Modulus√(x² + y²), positive
Argumentfrom positive real axis
Below the axisnegative argument
Multiply× moduli, + arguments
Divide÷ moduli, − arguments
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • No calculator: sketch every point first.
  • Give question 3 exactly.
GivenFind
1−√3 + imodulus, argument
2−3 − 3ipolar form
36 cis(−π/3)a + bi, exactly
42cis(3π/4) × 3cis(π/2)polar form
Show answer

1. Second quadrant: modulus 2, argument π − π/6 = 5π/6. 2. Third quadrant: 3√2 cis(−3π/4), not 3π/4 or π/4. 3. 6(1/2) + 6(−√3/2)i = 3 − 3√3i. 4. 6 cis(5π/4), brought into range as 6 cis(−3π/4).

Worked example

The answers: questions 1 and 2

  1. 1. −√3 + i: second quadrant
  2. r = √(3 + 1) = √4 = 2
  3. reference: tan⁻¹(1/√3) = π/6
  4. θ = π − π/6 = 5π/6
  5. 2. −3 − 3i: third quadrant
  6. r = √(9 + 9) = √18 = 3√2
  7. reference: tan⁻¹(3/3) = π/4
  8. θ = −(π − π/4) = −3π/4below the axis
  9. −3 − 3i = 3√2 cis(−3π/4)
Worked example

The answers: questions 3 and 4

  1. 3. x = 6 cos(−π/3) = 6(1/2) = 3
  2. y = 6 sin(−π/3) = 6(−√3/2) = −3√3
  3. 6 cis(−π/3) = 3 − 3√3i
  4. 4. modulus 2 × 3 = 6multiply moduli
  5. argument 3π/4 + π/2 = 5π/4add arguments
  6. 5π/4 − 2π = −3π/4more than π
  7. 6 cis(−3π/4)

Is your child on track?

Take the free placement test