Solutions — Operations with Complex Numbers

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1. Q1 — Addition and Subtraction

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   Add/subtract real parts, then imaginary parts separately.

   (a) (4 + 3i) + (2 − 5i) = (4 + 2) + (3 − 5)i = 6 − 2i

   (b) (4 + 3i) − (2 − 5i) = (4 − 2) + (3 − (−5))i = 2 + 8i

   (c) 2z = 2(4 + 3i) = 8 + 6i  →  2z + w = (8 + 6i) + (2 − 5i) = 10 + i

2. Q2 — Multiplying Complex Numbers

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   Expand like brackets, then replace i² = −1 and collect.

   (a) (2 + 3i)(1 + i) = 2(1) + 2(i) + 3i(1) + 3i(i) = 2 + 2i + 3i + 3i² = 2 + 5i − 3 = −1 + 5i

   (b) (3 − i)(3 + i) = 3² − i² = 9 − (−1) = 10. This is a conjugate pair — always gives a real result.

   (c) i(4 − 2i) = 4i − 2i² = 4i − 2(−1) = 2 + 4i

3. Q3 — Dividing Complex Numbers

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   Multiply numerator and denominator by the conjugate of the denominator.

   (a) 1/(1 + i) × (1 − i)/(1 − i):

   Numerator: 1 × (1 − i) = 1 − i

   Denominator: (1 + i)(1 − i) = 1 + 1 = 2

   Result: (1 − i)/2 = 1/2 − i/2

   (b) (2 + i)/(3 − i) × (3 + i)/(3 + i):

   Numerator: (2 + i)(3 + i) = 6 + 2i + 3i + i² = 6 + 5i − 1 = 5 + 5i

   Denominator: (3 − i)(3 + i) = 9 + 1 = 10

   Result: (5 + 5i)/10 = 1/2 + i/2

4. Q4 — Scalar Multiplication

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   Multiply the real scalar into both the real and imaginary parts.

   (a) 3(−2 + 5i) = −6 + 15i

   (b) −2(−2 + 5i) = 4 − 10i

   (c) (1/2)(−2 + 5i) = −1 + 5i/2

5. Q5 — Combined Operations

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   z = 1 + 2i, w = 3 − i, z̅ = 1 − 2i

   (a) z² = (1 + 2i)² = 1 + 4i + 4i² = 1 + 4i − 4 = −3 + 4i

   (b) zw = (1 + 2i)(3 − i) = 3 − i + 6i − 2i² = 3 + 5i + 2 = 5 + 5i

   zw − z̅ = (5 + 5i) − (1 − 2i) = 4 + 7i

   (c) z + w = (1 + 3) + (2 − 1)i = 4 + i

   z − w = (1 − 3) + (2 + 1)i = −2 + 3i

   (z + w)(z − w) = (4 + i)(−2 + 3i) = −8 + 12i − 2i + 3i² = −8 + 10i − 3 = −11 + 10i

6. Q6 — Division with Result Verification

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   (5 + 3i)/(2 + i) × (2 − i)/(2 − i):

   Numerator: (5 + 3i)(2 − i) = 10 − 5i + 6i − 3i² = 10 + i + 3 = 13 + i

   Denominator: (2 + i)(2 − i) = 4 + 1 = 5

   Result: (13 + i)/5 = 13/5 + i/5

   Verification: Multiply answer by (2 + i):

   (13/5 + i/5)(2 + i) = 26/5 + 13i/5 + 2i/5 + i²/5 = 26/5 + 15i/5 − 1/5 = 25/5 + 15i/5 = 5 + 3i ✓

7. Q7 — Solving for Unknown Complex Number

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   2z + 3i = (1 + i)z + (4 − i)

   Expand right side: 2z + 3i = z + iz + 4 − i

   Move all z terms left: 2z − z − iz = 4 − i − 3i

   z(1 − i) = 4 − 4i = 4(1 − i)

   Divide by (1 − i): z = 4(1 − i)/(1 − i) = 4

   Check: 2(4) + 3i = 8 + 3i. Right side: (1 + i)(4) + (4 − i) = 4 + 4i + 4 − i = 8 + 3i. ✓

8. Q8 — Real and Imaginary Parts After Operations

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   Multiply numerator and denominator by (1 − bi):

   Numerator: (2 + ai)(1 − bi) = 2 − 2bi + ai − abi² = (2 + ab) + (a − 2b)i

   Denominator: (1 + bi)(1 − bi) = 1 + b²

   So z = (2 + ab)/(1 + b²) + [(a − 2b)/(1 + b²)]i

   For z to be real, the imaginary part equals zero:

   (a − 2b)/(1 + b²) = 0 → a − 2b = 0 → a = 2b

9. Q9 — Simplifying a Complex Fraction

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   First simplify (1 + i)/(1 − i).

   Multiply by (1 + i)/(1 + i): (1 + i)² / ((1)² + (1)²) = (1 + 2i + i²)/2 = (1 + 2i − 1)/2 = 2i/2 = i

   So [(1 + i)/(1 − i)]² = i² = −1 (which is −1 + 0i)

   Alternative: compute (1 + i)² = 2i and (1 − i)² = −2i directly, then 2i/(−2i) = −1.

10. Q10 — Complex Equation with Two Unknowns

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    Let p + qi be a square root of 5 + 12i, where p, q ∈ ℝ.

    (p + qi)² = p² − q² + 2pqi = 5 + 12i

    Equate parts: p² − q² = 5 …(1)    and    2pq = 12, so pq = 6 …(2)

    From (2): q = 6/p. Substitute into (1):

    p² − 36/p² = 5  →  multiply by p²: p&sup4; − 5p² − 36 = 0

    Let u = p²: u² − 5u − 36 = 0 → (u − 9)(u + 4) = 0

    Since p is real, p² ≥ 0, so u = 9 → p = ±3.

    p = 3: q = 6/3 = 2  →  3 + 2i

    p = −3: q = 6/(−3) = −2  →  −3 − 2i

    Both values are valid; they are negatives of each other (as expected for square roots).