Mathematics 10 — Exercise Review · Q7

Full question

Simplify the following expression and find its real and imaginary parts.

Step-by-step solution

1. A negative exponent asks for a reciprocal. Find the reciprocal first.

   $$\frac{1}{7\times i+2}=\frac{1}{7\times i+2}$$

2. Start with the original quotient. For an inverse, the numerator is 1.

   $$\frac{1}{7\times i+2}$$

3. Find the conjugate of the denominator: keep its real part and reverse the imaginary sign.

   $$2-7\times i$$

4. Multiply by the conjugate over itself; this fraction equals 1.

   $$\left(\frac{2-7\times i}{2-7\times i}\right)\times \left(\frac{1}{7\times i+2}\right)$$

5. Multiply numerators together and denominators together. Keep their brackets intact.

   $$\frac{2-7\times i}{\left(2-7\times i\right)\times \left(7\times i+2\right)}$$

6. Calculate the numerator first.

7. Multiplication by 1 leaves the conjugate unchanged.

   $$2-7\times i=2-7\times i$$

8. Now calculate the denominator independently.

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

   $$7\times i\times \left(2-7\times i\right)+2\times \left(2-7\times i\right)$$

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

    $$2\times 2+2\times \left(-1\right)\times 7\times i+7\times 2\times i+7\times \left(-7\right)\times ii$$

11. Real × real: multiply the numerical coefficients.

    $$2\times 2=4$$

12. Real × imaginary: multiply the numerical coefficients.

    $$2\times \left(-1\right)\times 7\times i=-14\times i$$

13. Imaginary × real: multiply the numerical coefficients.

    $$7\times 2\times i=14\times i$$

14. Imaginary × imaginary: multiply the numerical coefficients.

    $$7\times \left(-7\right)\times ii=-49\times i^2$$

15. Substitute these four products into the expansion.

    $$4-14\times i+14\times i-49\times i^2$$

16. In the last term replace i² by −1.

    $$-49\times i^2=\left(-49\right)\times \left(-1\right)$$

17. Multiply its coefficient by −1.

    $$\left(-49\right)\times \left(-1\right)=49$$

18. Rewrite the expansion after this sign calculation.

    $$4-14\times i+14\times i+49$$

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

    $$4+49+i\times \left(-14+14\right)$$

20. Calculate the sum of the real terms.

    $$4+49=53$$

21. Calculate the sum of the imaginary coefficients.

    $$-14+14=0$$

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

    $$0\times i+53$$

23. The imaginary coefficient in the denominator is zero, so the denominator is real.

    $$53=53$$

24. Put the calculated numerator above the calculated denominator.

    $$\frac{2-7\times i}{53}$$

25. Separate the real term and imaginary term into two fractions with the same denominator.

    $$\frac{2}{53}+\frac{-7\times i}{53}$$

26. Move i outside its numerical fraction.

    $$\frac{2}{53}+\left(-\frac{7}{53}\right)\times i$$

27. Simplify the real coefficient fraction.

    $$\frac{2}{53}=\frac{2}{53}$$

28. For multiplication of fractions, multiply the numerators together and the denominators together.

    $$\frac{\left(-7\right)\times 1}{1\times 53}$$

29. Calculate these two products separately.

    $$-\frac{7}{53}$$

30. Calculate this operation before combining the remaining terms.

    $$\left(-7\right)\times \frac{1}{53}=-\frac{7}{53}$$

31. Simplify the imaginary coefficient fraction.

    $$-\frac{7}{53}=-\frac{7}{53}$$

32. Combine the reduced real and imaginary terms.

    $$\frac{2}{53}-\frac{7\times i}{53}$$

33. Now read the constant term as Re(z) and the coefficient of i as Im(z).

    $$(\mathrm{Re}=\frac{2}{53},\mathrm{Im}=-\frac{7}{53})$$

Answer

The final answer is shown below.