How does $1/i =-1$ where $i$ is an imaginary number?

Question

I was looking through this... and the math didn't work out. How does it work out that $1/i$ is equal to $-1$? I have tried working it out, but it equals $i/(-1)$ when I try to simplify.

Answer 1

You're right, and the book is, unfortunately, wrong.

Note that $i^{-1}$ is not $-1$, it is $-i$.

This follows as $$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=i \times (-1)=-i\neq -1$$

Answer 2

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

Answer 3

It doesn't work, because it's not true! You are correct that $$\frac{1}{i}=\frac{1}{i}\cdot\frac{i}{i}=\frac{i}{-1}=-i,$$ which is not $-1$!

Another way to see this is that $i\cdot (-i)=-i^2=-(-1)=1$, so $1/i=-i$.

Answer 4

By definition, $\frac{1}{i}$ refers to the multiplicative inverse of $i$. That is it gives $1$ when multiplied by $i$.

Note that $i\times -i = -(i\times i) = -(-1) = 1$.

Therefore, by definition, the multiplicative inverse of $i$ is $-i$,

That is, $\frac{1}{i} = -i$.