Mathematics 10 — Exercise Review · Q3 (iii)

Full question

Simplify the following expression.

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.

   $$\frac{1}{{\left(-i\right)}^9}=\frac{1}{-i^9}$$

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

   $$\frac{1}{i^9}=\frac{1}{i^9}$$

6. Divide 9 by 4: 9 = 4 × 2 + 1.

   $$i^9=i^1\times {\left(i^4\right)}^2$$

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

   $$i^9=1^2\times i$$

8. Take the reciprocal of this result.

   $$\frac{1}{i^9}=\frac{1}{i}$$

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

   $$\frac{1}{i}=\frac{i}{ii}$$

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

    $$\frac{1}{{\left(-i\right)}^9}=--1\times i$$

11. The simplified value is therefore:

    $$\frac{1}{{\left(-i\right)}^9}=i$$

Answer

The final answer is shown below.