Simultaneous equations with complex coefficients

Same method, stranger numbers

You have solved pairs like x + y = 5 and x − y = 1 since Class 8. Now the numbers in front of z and w can be complex, like (3 + i) or 2i. Does the old method still work?

It does. Simultaneous equations must be satisfied by the same values of the variables. Substitution works exactly as before; only the arithmetic uses i.

Solve complex simultaneous equations by substitution, exactly as with real ones, using i² = −1 in the arithmetic.

Key terms

Simultaneous equations

A system of equations solved together.

Complex coefficient

A complex number multiplying a variable.

Substitution

Replace one variable by its expression.

Check

Put z and w back into both equations.

Short questions with model answers

1. Q1. Example 21 (p. 19): 5z − (3 + i)w = 7 − i and (2 − i)z + 2iw = 4.

   • w = (5z − 7 + i) ÷ (3 + i)
   • (2 − i)(3 + i)z + 2i(5z − 7 + i) = 4(3 + i)
   • (7 + 9i)z − 2 − 14i = 12 + 4i, so (7 + 9i)z = 2(7 + 9i)
   • z = 2, w = (10 − 7 + i) ÷ (3 + i) = 1

   z = 2, w = 1

2. Q2. Exercise 1.4 Q2(iii) (p. 20): z − 4w = 3i and 2z + 3w = 11 − 5i.

   • z = 3i + 4w
   • 2(3i + 4w) + 3w = 11 − 5i, so 11w = 11 − 11i
   • w = 1 − i, z = 3i + 4 − 4i

   z = 4 − i, w = 1 − i

Common mistakes

• ✗ Multiply only the fraction by 3 + i.

  ✓ Every term on both sides gets multiplied, including 4 on the right.

• ✗ 2i(5z − 7 + i) = 10iz − 14i + 2i.

  ✓ 2i × i = 2i² = −2, so the last term is −2.

• ✗ Stop as soon as you find z.

  ✓ Find w too, and check both equations.

MCQs

1. 1. In Example 21, w =

   1. (a) 2
   2. (b) 1
   3. (c) 3 + i
   4. (d) i
   Show answer

   (b) (5 × 2 − 7 + i) ÷ (3 + i) = (3 + i) ÷ (3 + i) = 1.

2. 2. Simultaneous equations are equations that:

   1. (a) Have only real coefficients
   2. (b) Each have a different solution
   3. (c) Are satisfied by the same values of the variables
   4. (d) Have no solution
   Show answer

   (c) Definition on p. 19.

3. 3. 2i × i equals:

   1. (a) −2
   2. (b) 2
   3. (c) 2i
   4. (d) −2i
   Show answer

   (a) 2i² = 2(−1) = −2.

4. 4. z + w = 3i and 2z + 3w = 2. Find w (Exercise 1.4 Q2(iv)).

   1. (a) −2 + 9i
   2. (b) 2 + 6i
   3. (c) 1 − i
   4. (d) 2 − 6i
   Show answer

   (d) 2(3i − w) + 3w = 2 gives w = 2 − 6i.

5. 5. (2 − i)(3 + i) =

   1. (a) 5 − i
   2. (b) 7 − i
   3. (c) 6 − i
   4. (d) 7 + i
   Show answer

   (b) 6 + 2i − 3i − i² = 6 − i + 1 (p. 19).

Quick revision

• Make one variable the subject, substitute, then back-substitute (Steps I–III).
• Complex coefficients only change the arithmetic: use i² = −1.
• Always check z and w in both equations.