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

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