Mathematics 10 — Exercise 1.2 · Q1 (v)

Full question

Simplify the following expression and express the result in the standard form a + bi, where a and b are real numbers.

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.

   $$4\times i\times \left(2-3\times i\right)+3\times \left(2-3\times i\right)$$

2. Distribute each of those products. Keep all four multiplications visible.

   $$3\times 2+3\times \left(-1\right)\times 3\times i+4\times 2\times i+4\times \left(-3\right)\times ii$$

3. Real × real: multiply the numerical coefficients.

   $$3\times 2=6$$

4. Real × imaginary: multiply the numerical coefficients.

   $$3\times \left(-1\right)\times 3\times i=-9\times i$$

5. Imaginary × real: multiply the numerical coefficients.

   $$4\times 2\times i=8\times i$$

6. Imaginary × imaginary: multiply the numerical coefficients.

   $$4\times \left(-3\right)\times ii=-12\times i^2$$

7. Substitute these four products into the expansion.

   $$6-9\times i+8\times i-12\times i^2$$

8. In the last term replace i² by −1.

   $$-12\times i^2=\left(-12\right)\times \left(-1\right)$$

9. Multiply its coefficient by −1.

   $$\left(-12\right)\times \left(-1\right)=12$$

10. Rewrite the expansion after this sign calculation.

    $$6-9\times i+8\times i+12$$

11. Group the real terms and factor i out of the imaginary terms.

    $$6+12+i\times \left(-9+8\right)$$

12. Calculate the sum of the real terms.

    $$6+12=18$$

13. Calculate the sum of the imaginary coefficients.

    $$-9+8=-1$$

14. Insert both calculated sums in a + bi form.

    $$18-i$$

Answer

The final answer is shown below.