Question

Can you help me understand how to only divide by i? Example: 14+i/i

Answer 1

Dividing by `i` is the same as multiplying by `-i`.

Explanation:

Assuming you mean `\frac{14+i}{i}`, otherwise `14+i/i = 14+1=15`

But anyway, since you ask it, let's see how to divide by `i` in general: you can perform some sort of rationalization, multiplying and dividing by `i`:

`\frac{14+i}{i}*\frac{i}{i} = \frac{i(14+i)}{i^2}= \frac{i(14+i)}{-1} = -i(14+i)`

So, dividing by `i` is the same as multiplying by `-i`.

This makes particularly sense if you notice the periodicity of the powers of `i`:

`i^0 = 1`
`i^1 = i`
`i^2 = -1`
`i^3 = -i`
`i^4 = 1`
`i^5 = i`
`...`

In general, you have `i^{4n+k} = i^k`. But this is true for negative exponents as well!

`i^{-4} = 1`
`i^{-3} = i`
`i^{-2} = -1`
`i^{-1} = -i`
`i^0 = i`
`...`

In fact, you can write

`1/i = i^-1 = i^{4*(-1)+3} = i^3 = -i`

And this is why dividing by `i` is like multiplying by `-i`

Answer 2

You multiply by `1` in the from `i/i`; this makes the divisor become `-1` because `i xx i = -1`

Explanation:

Given: `(14+i)/i`

Multiply by `1` in the form of `i/i`

`(14+i)/i i/i`

The denominator becomes `-1`:

`(i(14+i))/-1`

Use the distributive property to multiply each term in the denominator by `i`

`(14i+i^2)/-1`

We know that `i^2 = -1`:

`(14i-1)/-1`

Divide by -1:

`1 -14i`