Complex arithmetic: 7 exam questions solved step by step

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What this lesson covers

  • A complex number has a real part and an imaginary part, and equating those parts turns one equation into two.
  • Multiplying by the conjugate is how a complex denominator, or a surd denominator, is made real.
Key idea

7 questions, one toolkit

  • 1. Expand, then i² = −1.
  • 2. Divide: × the bottom's conjugate.
  • 3. Equal numbers: equate both parts.
  • 4. Answer in the form asked.
Your turn

Question 1 · Achieved

  • z = 4 − 3i and w = 2 + 5i
  • Find 2z − w and zw in the form a + bi.
Worked example

Q1: 2z − w and zw

  • Add: group the parts.
  • Multiply: expand, i² = −1.
  1. 2z = 8 − 6i
  2. 2z − w = (8 − 2) + (−6 − 5)igroup the parts
  3. = 6 − 11i
  4. zw = (4 − 3i)(2 + 5i)
  5. = 8 + 20i − 6i − 15i²expand
  6. = 8 + 14i + 15i² = −1
  7. = 23 + 14i
Your turn

Question 2 · Achieved

  • Write (2 + √3)/(2 − √3)
  • in the form a + b√3.
Worked example

Q2: (2 + √3)/(2 − √3)

  • Conjugate of the bottom.
  • Square the top fully.
  1. × (2 + √3)/(2 + √3)conjugate of the bottom
  2. bottom: (2 − √3)(2 + √3)
  3. = 4 − 3 = 1√3 × √3 = 3
  4. top: (2 + √3)² = 4 + 4√3 + 3
  5. = 7 + 4√3
  6. (7 + 4√3)/1 = 7 + 4√3
  7. (2 − √3)(7 + 4√3) = 2 + √3
Your turn

Question 3 · Merit

  • Write (4 + 7i)/(3 − 2i)
  • in the form a + bi.
Worked example

Q3: (4 + 7i)/(3 − 2i)

  • The bottom becomes real.
  • Then split the fraction.
  1. × (3 + 2i)/(3 + 2i)conjugate of the bottom
  2. bottom: 9 + 4 = 13a² + b²
  3. top: (4 + 7i)(3 + 2i)
  4. = 12 + 8i + 21i + 14i²expand
  5. = −2 + 29ii² = −1
  6. (−2 + 29i)/13not finished
  7. = −2/13 + (29/13)ithe form asked
  8. (3 − 2i)(−2 + 29i) = 52 + 91i
  9. (52 + 91i)/13 = 4 + 7i
Your turn

Question 4 · Merit

  • k is real, and (k + 3i)/(2 − i)
  • is purely imaginary.
  • Find k, and the number.
Worked example

Q4: make the bottom real

  • Same conjugate method.
  • Group with k inside.
  1. × (2 + i)/(2 + i)conjugate of the bottom
  2. bottom: 4 + 1 = 5
  3. top: (k + 3i)(2 + i)
  4. = 2k + ki + 6i + 3i²expand
  5. = (2k − 3) + (k + 6)ii² = −1, group
  6. ((2k − 3) + (k + 6)i)/5
Worked example

Q4: purely imaginary

  • Real part = 0.
  • Not the imaginary part.
  1. ((2k − 3) + (k + 6)i)/5
  2. 2k − 3 = 0real part is 0
  3. k = 3/2
  4. k + 6 = 0that makes it real
  5. imaginary: (3/2 + 6) ÷ 5
  6. = (15/2) ÷ 5 = 3/2
  7. the number is (3/2)i
  8. (3/2)i × (2 − i) = 3/2 + 3i
Your turn

Question 5 · Merit, in context

  • AC circuit, two parts in parallel:
  • Z₁ = 2 + 4i Ω, Z₂ = 4 − 2i Ω
  • Total impedance Z = Z₁Z₂ ÷ (Z₁ + Z₂)
  • Find Z in the form a + bi.
Worked example

Q5: top and bottom

  • Product, then sum.
  • Then one division.
  1. Z₁Z₂ = (2 + 4i)(4 − 2i)
  2. = 8 − 4i + 16i − 8i²expand
  3. = 16 + 12ii² = −1
  4. Z₁ + Z₂ = 6 + 2iadd the parts
  5. Z = (16 + 12i)/(6 + 2i)
Worked example

Q5: divide

  • Conjugate of 6 + 2i.
  • Answer with units.
  1. Z = (16 + 12i)/(6 + 2i)
  2. × (6 − 2i)/(6 − 2i)conjugate of the bottom
  3. bottom: 36 + 4 = 40
  4. top: 96 − 32i + 72i − 24i²expand
  5. = 120 + 40ii² = −1
  6. (120 + 40i)/40 = 3 + idivide both parts
  7. total impedance 3 + i ohms
Your turn

Question 6 · Excellence

  • Solve z² = −5 + 12i
  • Give both answers in the form a + bi.
  • No calculator: show the algebra.
Worked example

Q6: equate the parts

  • Let z = x + iy.
  • One equation becomes two.
  1. z = x + iy, x and y real
  2. (x + iy)² = −5 + 12i
  3. x² + 2xyi + i²y² = −5 + 12iexpand
  4. (x² − y²) + 2xyi = −5 + 12ii² = −1
  5. x² − y² = −5real parts
  6. 2xy = 12, so y = 6/ximaginary parts
Worked example

Q6: solve, reject, check

  • A quadratic in x².
  • x is real: reject x² < 0.
  1. x² − y² = −5, y = 6/x
  2. x² − 36/x² = −5substitute y = 6/x
  3. x⁴ + 5x² − 36 = 0multiply by x²
  4. (x² + 9)(x² − 4) = 0factorise
  5. x² = −9x is real
  6. x² = 4, x = 2 or x = −2
  7. x = 2: y = 3; x = −2: y = −3
  8. z = 2 + 3i or z = −2 − 3i
  9. (2 + 3i)² = 4 + 12i − 9 = −5 + 12i
Your turn

Question 7 · Excellence

  • Solve |z|² + 2z = 12 − 4i
  • Give every solution in the form a + bi.
Worked example

Q7: write z as x + iy

  • |z|² is real.
  • Imaginary part first.
  1. z = x + iy, x and y real
  2. |z|² = x² + y²modulus squared
  3. x² + y² + 2x + 2yi = 12 − 4i
  4. 2y = −4, so y = −2imaginary parts
  5. x² + y² + 2x = 12real parts
  6. x² + 4 + 2x = 12y = −2
  7. x² + 2x − 8 = 0
Worked example

Q7: two solutions

  • x only has to be real.
  • Check both in the original.
  1. x² + 2x − 8 = 0, y = −2
  2. (x + 4)(x − 2) = 0factorise
  3. x = −4 or x = 2both real: keep both
  4. z = 2 − 2i: 8 + (4 − 4i)
  5. = 12 − 4i
  6. z = −4 − 2i: 20 + (−8 − 4i)
  7. = 12 − 4i
  8. z = 2 − 2i or z = −4 − 2i
Summary

The toolkit that solved all seven

  • Expand, use the conjugate, equate both parts, and check.
StepHow
Expandthen i² = −1 at once
Divide× the bottom’s conjugate
Surdsthe same conjugate trick
Unknownsequate real, imaginary
Answera + bi, the form asked
Checkmultiply back or square

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