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.
How does $1/i =-1$ where $i$ is an imaginary number?
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$\begingroup$
complex-numbers
arithmetic
quadratics
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1thank you, this was very helpful – 2017-02-24
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01/i doesn't equal -1. 1/i = -i. Because i (-i)=-i^2=-(-1)=1. – 2017-02-24
4 Answers
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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 $1/i=-i$.
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$\frac{1}{i} = \frac{i}{i^{2}} = \frac{i}{-1} = -i$
3
You're right, and the book is, unfortunately, wrong.
Note that $i^{-1}$ is not $-1$, it is $-i$.
This follows as $$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=i \times (-1)=-i\neq -1$$
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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, $\frac{1}{i} = -i$.