Mathematics 10 — Exercise Review · Q6 (iv)

Full question

For z₁ = 5 + 4i, z₂ = 3 + 2i, evaluate |z₁z₂|.

(2×i+3)×(4×i+5)

Step-by-step solution

  1. Distribute the first bracket: multiply its real term by the whole second bracket, then its imaginary term by that bracket.

    4×i×(2×i+3)+5×(2×i+3)
  2. Distribute each of those products. Keep all four multiplications visible.

    5×3+5×2×i+4×3×i+4×2×ii
  3. Real × real: multiply the numerical coefficients.

    5×3=15
  4. Real × imaginary: multiply the numerical coefficients.

    5×2×i=10×i
  5. Imaginary × real: multiply the numerical coefficients.

    4×3×i=12×i
  6. Imaginary × imaginary: multiply the numerical coefficients.

    4×2×ii=8×i2
  7. Substitute these four products into the expansion.

    15+10×i+12×i+8×i2
  8. In the last term replace i² by −1.

    8×i2=8×(-1)
  9. Multiply its coefficient by −1.

    8×(-1)=-8
  10. Rewrite the expansion after this sign calculation.

    15+10×i+12×i−8
  11. Group the real terms and factor i out of the imaginary terms.

    15−8+i×(10+12)
  12. Calculate the sum of the real terms.

    15−8=7
  13. Calculate the sum of the imaginary coefficients.

    10+12=22
  14. Insert both calculated sums in a + bi form.

    22×i+7
  15. Read a, the real coordinate.

    a=7
  16. Read b, the coefficient of i, including its sign.

    b=22
  17. Use |z| = √(a² + b²) and substitute both coordinates.

    72+222
  18. Square the real coordinate.

    72=49
  19. Square the imaginary coordinate; a negative number squared is positive.

    222=484
  20. Add the two squared values.

    49+484=533
  21. Take the nonnegative square root of this sum.

    533=533

Answer

The final answer is shown below.

533