# Integral Powers of Iota – Complex Numbers

Here you will learn complex number i (iota) and integral powers of iota with examples.

Let’s begin –

## Integral Powers of Iota (i)

Positive Integral Powers of iota(i) :

We have,

i = $$\sqrt{-1}$$

$$\therefore$$ $$i^2$$ = -1

$$i^3$$ = $$i^2$$ $$\times$$ i = -i

$$i^4$$ = $$(i^2)^2$$ = $$(-1)^2$$ = 1

In order to compute $$i^n$$ for n > 4, we divide n by 4 and obtain the remainder r. Let m be the quotient when n is divided by 4. Then,

n = 4m + r, where 0 $$\le$$ r < 4

$$\implies$$ $$i^n$$ = $$i^{4m + r}$$ = $$(i^4)^m i^r$$ = $$i^4$$

Thus, the value of $$i^n$$ for n > 4 is $$i^r$$, where r is the remainder when n is divide by 4.

Negative integral powers of iota(i) :

By the law of indices, we have

$$i^{-1}$$ = $$1\over i$$ = $$i^3\over i^4$$ = $$i^3$$ = -i

$$i^{-2}$$ = $$1\over i^2$$ = $$1\over -1$$ = -1

$$i^{-3}$$ = $$1\over i^3$$ = $$1\over i^4$$ = i

$$i^{-4}$$ = $$1\over i^4$$ = $$1\over 1$$ = 1

If n > 4, then

$$i^{-n}$$ = $$1\over i^n$$ = $$1\over i^r$$, where r is the remainder when n is divided by 4.

Note : $$i^{0}$$ is defined as 1.

Example : Evaluate the following :

(i) $$i^{135}$$

(ii) $$i^{19}$$

(iii) $$i^{-999}$$

Solution

(i) 135 leaves remainder as 3 when it is divided by 4.

$$\therefore$$  $$i^{135}$$ = $$i^3$$ = -i

(ii) The remainder is when 19 is divided by 4.

$$\therefore$$  $$i^{19}$$ = $$i^3$$ = -i

(iii) We have, $$i^{-999}$$ = $$1\over i^{999}$$

On dividing 999 by 4, we obtain 3 as the remainder.

$$\therefore$$ $$i^{999}$$ = $$i^3$$

$$\implies$$ $$i^{-999}$$ = $$1\over i^{999}$$ = $$1\over i^3$$ = $$i\over i^4$$ = $$i\over 1$$ = i