Mathematics 10 — Exercise 1.1 · Q1 (iii)

Full question

Simplify the following power of iota, using i² = −1.

Step-by-step solution

1. By definition, the square of iota is −1.

   $$i^2=-1$$

2. Multiply i² by i to obtain i³.

   $$i^3=-i$$

3. Square i² to obtain i⁴: (−1) × (−1) = 1.

   $$i^4={\left(-1\right)}^2$$

4. Separate the sign from the power of i.

   $$\frac{1}{{\left(-i\right)}^{19}}=\frac{1}{-i^{19}}$$

5. First evaluate the positive power; a negative power is its reciprocal.

   $$\frac{1}{i^{19}}=\frac{1}{i^{19}}$$

6. Divide 19 by 4: 19 = 4 × 4 + 3.

   $$i^{19}=i^3\times {\left(i^4\right)}^4$$

7. Every i⁴ contributes 1. Substitute the value of the remaining power.

   $$i^{19}=1^4\times \left(-i\right)$$

8. Take the reciprocal of this result.

   $$\frac{1}{i^{19}}=\frac{1}{-i}$$

9. Multiply numerator and denominator by i. Replace the denominator i² with −1, then simplify the signs.

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

10. An odd power of −1 is −1; an even power is 1. Multiply this sign by the value just found.

    $$\frac{1}{{\left(-i\right)}^{19}}=-i$$

11. The simplified value is therefore:

    $$\frac{1}{{\left(-i\right)}^{19}}=-i$$

Answer

The final answer is shown below.