Polar form: 7 exam questions solved step by step

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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

7 questions, one method

  • 1. Sketch it: name the quadrant.
  • 2. r = √(x² + y²), never negative.
  • 3. Reference angle, then adjust.
  • 4. Keep −π < θ ≤ π.
Your turn

Question 1 · Achieved

  • Write 8 cis(−5π/6)
  • in the form a + bi, exactly.
Worked example

Q1: 8 cis(−5π/6)

  • x = r cos θ, y = r sin θ.
  • Exact values only.
  1. x = r cos θ, y = r sin θ
  2. −5π/6: third quadrantreference π/6
  3. cos(−5π/6) = −√3/2
  4. sin(−5π/6) = −1/2
  5. x = 8(−√3/2) = −4√3
  6. y = 8(−1/2) = −4
  7. 8 cis(−5π/6) = −4√3 − 4i
  8. √(48 + 16) = √64 = 8modulus
Your turn

Question 2 · Achieved

  • Write z = 3 + 3√3i
  • in polar form, exactly.
Worked example

Q2: z = 3 + 3√3i

  • Quadrant first.
  • Check by converting back.
  1. x > 0, y > 0first quadrant
  2. r = √(3² + (3√3)²)modulus
  3. = √(9 + 27) = √36 = 6
  4. θ = tan⁻¹(3√3/3) = tan⁻¹(√3)
  5. = π/3first quadrant: keep
  6. z = 6 cis(π/3)
  7. 6(1/2) + 6(√3/2)i = 3 + 3√3i
Your turn

Question 3 · Merit

  • Write w = −√2 − √6i
  • in polar form, exactly.
  • Careful: sketch it before tan⁻¹.
Question 3 · Merit
Worked example

Q3: w = −√2 − √6i

  • The calculator gives π/3.
  • The sketch says no.
  1. x < 0, y < 0third quadrant
  2. r = √((−√2)² + (−√6)²)modulus
  3. = √(2 + 6) = √8 = 2√2
  4. tan⁻¹(−√6/−√2) = π/3first quadrant
  5. reference angle π/3
  6. θ = −(π − π/3) = −2π/3below the real axis
  7. w = 2√2 cis(−2π/3)
  8. 2√2(−1/2) = −√2, 2√2(−√3/2) = −√6
Your turn

Question 4 · Merit

  • u = 4 cis(2π/3), v = 2 cis(−3π/4)
  • Find uv and u/v in polar form,
  • with −π < θ ≤ π.
Worked example

Q4: uv and u/v

  • × moduli, + arguments.
  • ÷ moduli, − arguments.
  1. |uv| = 4 × 2 = 8multiply moduli
  2. arg = 2π/3 + (−3π/4)add arguments
  3. = 8π/12 − 9π/12 = −π/12in range
  4. uv = 8 cis(−π/12)
  5. |u/v| = 4 ÷ 2 = 2divide moduli
  6. arg = 2π/3 − (−3π/4)subtract arguments
  7. = 8π/12 + 9π/12 = 17π/12more than π
  8. 17π/12 − 24π/12 = −7π/12subtract 2π
  9. u/v = 2 cis(−7π/12)
Your turn

Question 5 · Merit, in context

  • A kayaker leaves a jetty. East is real,
  • north is imaginary, in km.
  • She is at −5√3 + 5i.
  • How far away is she, and which way?
Question 5 · Merit, in context
Worked example

Q5: distance and direction

  • Modulus: the distance.
  • Argument: the direction.
  1. x < 0, y > 0second quadrant
  2. r = √((−5√3)² + 5²)modulus
  3. = √(75 + 25) = √100 = 10
  4. reference: tan⁻¹(5/(5√3)) = π/6
  5. θ = π − π/6 = 5π/6second quadrant
  6. z = 10 cis(5π/6)
  7. 5π/6 = 150°: 30° north of west
  8. 10 km away, 30° north of west
Your turn

Question 6 · Excellence

  • z = (−1 + i)/(√3 + i)
  • a) Write z in polar form.
  • b) Write z in the form a + bi.
  • c) Hence find cos(7π/12) exactly.
Worked example

Q6 a): z in polar form

  • Top and bottom separately.
  • ÷ moduli, − arguments.
  1. −1 + i: r = √(1 + 1) = √2
  2. θ = π − π/4 = 3π/4second quadrant
  3. √3 + i: r = √(3 + 1) = 2
  4. θ = tan⁻¹(1/√3) = π/6first quadrant
  5. |z| = √2 ÷ 2 = √2/2divide moduli
  6. arg z = 3π/4 − π/6subtract arguments
  7. = 9π/12 − 2π/12 = 7π/12
  8. z = (√2/2) cis(7π/12)
Worked example

Q6 b): z in the form a + bi

  • Conjugate of the bottom.
  • Group, then split.
  1. × (√3 − i)/(√3 − i)conjugate of the bottom
  2. bottom: 3 + 1 = 4
  3. top: (−1 + i)(√3 − i)
  4. = −√3 + i + √3i − i²expand
  5. = (1 − √3) + (1 + √3)ii² = −1, group
  6. z = (1 − √3)/4 + ((1 + √3)/4)i
Worked example

Q6 c): cos(7π/12)

  • Same z, two forms.
  • Equate the real parts.
  1. z = (√2/2) cis(7π/12)
  2. z = (1 − √3)/4 + ((1 + √3)/4)i
  3. (√2/2) cos(7π/12) = (1 − √3)/4
  4. cos(7π/12) = (1 − √3)/(2√2)divide by √2/2
  5. = (√2 − √6)/4rationalise
  6. cos(7π/12) = (√2 − √6)/4
  7. 7π/12 > π/2, so cos < 0
  8. (√2 − √6)/4 ≈ −0.2588
Your turn

Question 7 · Excellence

  • w = (a + i)/(1 + i), where a is real.
  • arg w = π/3.
  • Find the exact value of a.
Worked example

Q7: w in the form x + iy

  • Conjugate of the bottom.
  • Keep a in each part.
  1. × (1 − i)/(1 − i)conjugate of the bottom
  2. bottom: 1 + 1 = 2
  3. top: (a + i)(1 − i)
  4. = a − ai + i − i²expand
  5. = (a + 1) + (1 − a)ii² = −1, group
  6. w = (a + 1)/2 + ((1 − a)/2)i
Worked example

Q7: use the argument

  • tan θ = y/x.
  • Check the quadrant.
  1. w = (a + 1)/2 + ((1 − a)/2)i
  2. arg w = π/3: x > 0, y > 0first quadrant
  3. tan(π/3) = y/x = (1 − a)/(a + 1)
  4. √3(a + 1) = 1 − atan(π/3) = √3
  5. √3a + a = 1 − √3collect a terms
  6. a = (1 − √3)/(1 + √3)
  7. = (4 − 2√3)/(−2)× (1 − √3) top, bottom
  8. = √3 − 2
  9. a + 1 > 0 and 1 − a > 0first quadrant
Summary

The method that solved all seven

  • Sketch, find r, find θ from the quadrant, keep it in range.
StepHow
Sketchname the quadrant first
Modulus√(x² + y²), never negative
Argumentreference angle, adjust
Below axisnegative argument
× and ÷× or ÷ r, + or − θ
Range−π < θ ≤ π

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