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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 $i$.

Community
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4yl1n
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  • Not exactly a duplicate, but an explanation of the paradox: https://math.stackexchange.com/questions/3345202/square-root-of-a-product-of-negative-numbers/3345233#3345233 – Ethan Bolker Dec 06 '19 at 00:54

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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!

J. W. Tanner
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Ty Jensen
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  • But $1/i$ is equal to $-i$ so there's not a lot wrong with the OP's point. You might find this page https://en.wikipedia.org/wiki/Branch_point helpful. – Rob Arthan Dec 06 '19 at 00:21
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    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. – Rob Arthan Dec 06 '19 at 00:44
  • 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. – Ty Jensen Dec 06 '19 at 01:05
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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, $\frac1i=-i.$

Kenta S
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Observe the following argument:

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

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

$$i\cdot i=-1$$

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

Andrew Chin
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R. Burton
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    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. – kimchi lover Dec 06 '19 at 01:19
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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 $\sqrt{-1}$ .

Rob Arthan
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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 $1/i=-i$.

Michael Hoppe
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