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When I first looked at the problem $\frac{1}{i}$ I reasoned: $$(\frac{1}{\sqrt{-1}})^2=(\frac{1}{\sqrt{-1}})(\frac{1}{\sqrt{-1}})=\frac{1}{-1}=-1$$ So $$(\frac{1}{\sqrt{-1}})^2=-1$$ And if you square root both sides of the equation you get $$\frac{1}{\sqrt{-1}}=\sqrt{-1}$$ Now I realize this reasoning from here: $$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$ But how can $\frac{1}{i}$ equal $i$ and $-i$ at the same time?

I understand that the answer will probable be: we really don't know much about $i$.

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Here, look at the function $f(x) = \sqrt{x}$ in Desmos: https://www.desmos.com/calculator/odlrlh9ozn

Notice anything? It is only defined over $[0, \infty)$. This also means that an identity of square roots like $$\sqrt{a}\sqrt{b} = \sqrt{ab}$$ is only defined for non-negative numbers. So when you are squaring $i$ you are essentially doing $$i^2 = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = 1$$ but also $$(-1^\frac{1}{2})^{2}=-1^{1}=-1$$ This "phenomenon" is not a phenomenon at all; it is a contradiction because negative numbers are not defined on the domain of the square root function. Imaginary numbers on not included in the set of all real numbers, so no wonder they don't act at all like real numbers!

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  • $\begingroup$ But $1/i$ is equal to $-i$ so there's not a lot wrong with the OP's point. You might find this page en.wikipedia.org/wiki/Branch_point helpful. $\endgroup$
    – Rob Arthan
    Commented Dec 6, 2019 at 0:21
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    $\begingroup$ I think you are doing the OP a disservice by not helping him or her with some of the standard facts about extending multi-valued functions like square root into the complex plane. The last two sentences of your answer make no sense at all. $\endgroup$
    – Rob Arthan
    Commented Dec 6, 2019 at 0:44
  • $\begingroup$ I really just wanted to demonstrate that i does not act like a real number (which it doesn't). I think trying to teach OP too much about the nature of the complex plane will just steamroll him/her and is out of scope of the question they are asking. $\endgroup$
    – Ty Jensen
    Commented Dec 6, 2019 at 1:05
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In general, when given the equation $x^2=a,$ there are two solutions (unless $a=0$): $x=\pm\sqrt a.$

In this case, both $i$ and $-i$ are solutions to the equation $x^2=-1.$

Thus, just because $(\frac1i)^2=-1,$ you cannot conclude $\frac1i=i,$ as there is the possibility that $\frac1i=-i.$ And in this case, as you said, $\frac1i=-i.$

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Hint: presumably, the question was to find the value of $\frac{1}{i}$. When working any field, if you want to test whether $\frac{1}{x} = y$, you should multiply through by $x$ and test whether $1=xy$. In $\Bbb{C}$, this method will indeed show you that $\frac{1}{i} = -i$ and that $\frac{1}{i} \neq i$ using only the assumption that $i^2 = -1$, without any dangerous algebraic manipulations involving $\sqrt{-1}$ .

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Observe the following argument:

$$\frac{1}{i}=i\implies \frac{1}{i}\cdot i=i\cdot i$$

$$\frac{1}{x}\cdot x=1\implies \frac{1}{i}\cdot i=1$$

$$i\cdot i=-1$$

$$\text{therefore, if } \frac{1}i=i\text{, then } 1=-1$$

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    $\begingroup$ Your answer would be better if (1) you pinpointed the OP's mistake, and (2) you mentioned up front that you are providing a reductio ad absurdum argument. $\endgroup$ Commented Dec 6, 2019 at 1:19
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Some of the rules of operations you learn in school, must be modified to accommodate the complex numbers:

1) $\forall k\in\mathbb{Z}; i=i^{4k+1}, -1=i^{4k+2},-i=i^{4k+3},1=i^{4k}$

Complex numbers do 2d rotation, when you multiply them.

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As multiplication by $i$ may be considered as counterclockwise rotation by $\pi/2$ and $\frac1i i=1$, we may consider $1/i$ as rotation clockwise by $\pi/2$, that is $1/i=-i$.

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