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