The value of `i^n` for various values of n is discussed.

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What is the value of `i`?

- `-1`
- `sqrt(-1)`
- `sqrt(-1)`

The answer is '`sqrt(-1)`'

What is the value of `i^2`?

- `-1`
- `-1`
- `sqrt(-1)`

The answer is '`-1`'

It is defined that `i=sqrt(-1)`. Substitute that in the following

`i^2`

`quad quad = (sqrt(-1))^2`

`quad quad = (-1^(1/2))^2`

`quad quad = -1^(2/2)`

`quad quad = -1`

What is the value of `i^3`?

- `-1`
- `i`
- `-i`
- `-i`

The answer is '`-i`'

`i^3`

`quad quad = i^(2+1)`

`quad quad = i^2 xx i^1`

`quad quad = -1 xx i`

`quad quad = -i`

What is the value of `i^4`?

- `1`
- `1`
- `-1`

The answer is '`1`'

`i^4`

`quad quad = i^(2+2)`

`quad quad = (i)^2 xx i^2`

`quad quad = -1 xx -1`

`quad quad = 1`

What is the value of `i^0`?

- `1`
- `1`
- `-1`
- `i`

The answer is '`1`'

`i^0`

`quad quad = i^(1-1)`

`quad quad = i/i`

`quad quad = 1`

What is the value of `i^n`, if `n=4p+2q+r`?

- `i`
- `(-1)^q xx i^r`
- `(-1)^q xx i^r`
- `i^r`

The answer is '`(-1)^q xx i^r`'

`i^n`

`quad quad = i^(4p+2q+r)`

`quad quad = i^(4p) xx i^(2q) xx i^r`

`quad quad = (i^4)^p) xx (i^2)^q) xx i^r`

`quad quad = (-1)^q xx i^r`

To calculate `i^n`, use `i^4=1` and `i^2=-1`.

**Power of i:** To calculate `i^n`, express `n` in the form `n=4p+2q+r`, where `p,q, r in NN` natural numbers. Then

`i^n = (-1)^q xx i^r`.

*Solved Exercise Problem: *

What is the value of `i^27`?

- `1`
- `-i`
- `-i`
- `i`

The answer is '`-i`'

*Solved Exercise Problem: *

What is `i^20`?

- `1`
- `1`
- `-1`
- `i`
- `-i`

The answer is '`1`'

*Solved Exercise Problem: *

What is `i^7`?

- `1`
- `-1`
- `i`
- `-i`
- `-i`

The answer is '`-i`'

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