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$$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \sqrt{-1} = i$$

I know this is wrong, but why? I often see people making simplifications such as $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$, and I would calculate such a simplification in the manner shown above, namely

$$\frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{\sqrt{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$

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marked as duplicate by Chris Culter, Asaf Karagila, Jyrki Lahtonen, Cameron Buie, Ayman Hourieh Dec 10 '13 at 20:24

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5  
Such simplifications only hold when you're dealing with real numbers. –  Alyosha Dec 9 '13 at 20:39
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In general $\sqrt{ab}=\sqrt a\sqrt b$ holds only when both $a,b>0$. –  Pedro Tamaroff Dec 9 '13 at 20:40
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$\sqrt{-1}$ is ambiguous : it could be $i$ or $-i$. –  Xoff Dec 9 '13 at 20:43
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IMHO the key slip here comes in saying that '$\sqrt{-1}=i$'. When working with reals we can often use order considerations (e.g. $\sqrt{x}\geq 0$) to eliminate any sign ambiguity when talking about $\pm\sqrt{x}$, but for complex numbers that ambiguity becomes much more inherent. –  Steven Stadnicki Dec 9 '13 at 20:44
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This has caused more of a stir than I hoped for! Huh, I always thought that $i$ was defined as $\sqrt{-1}$ but I guess it makes sense that there are always two values of a square root –  TomJS Dec 9 '13 at 20:52

9 Answers 9

up vote 43 down vote accepted

What you are doing is a version of $$ -1=i^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt1=1. $$ It simply shows that for non-positive numbers, it is not always true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$.

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6  
I must say I like OP's version a bit better than this because here it's true (in some sense) that $-1=\sqrt{1}$. Of course there's the same sort of ambiguity in talking about 'the' square root. –  Steven Stadnicki Dec 9 '13 at 20:43
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@StevenStadnicki It is certainly true that $-1 \in \{ x\in \mathbb{R} : x^2 = 1\}$. However, it is certainly not true that $-1 = \sqrt{1}$. For any real $x > 0$ we have, for any positive integer $r$, by definition, $y=\sqrt[r]{x}$ to be the unique positive real number for which $y^r=x$. –  Fly by Night Dec 9 '13 at 20:48

$$\frac1{\sqrt{-1}}=\sqrt{-1}$$ is only true in the sense that $1$ over a square root of $-1$ is a square root of $-1$. However, there are two square roots of every non-zero complex number, so you have to make sure to pick the right one. For square roots of non-negative real numbers it works to consistently pick the non-negative square root, but no such rule exists for all complex numbers.

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Even for real numbers, there are two square roots. $(-2)^2 = 2^2 =4$, so both $2$ and $-2$ can be thought of as square roots of $4$. In the real numbers, we have an easy way to pick one of the two: just always pick the positive one. So we define $\sqrt{x}$ to be the positive square root of $x$. In the complex case, we do not have an order, and so no consistent way to pick one of the two square roots. So $\sqrt{z}$ is not really a function: it is a multivalued function. You confusion arises from thinking that $\sqrt{i}$ indicates only one number.

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You assume $\sqrt{-1}=i$, but it's wrong to write such a thing, because the square root function is really defined as a function only for positive argument, and it has positive values. If you write everywhere $\pm \sqrt{a}$ instead of $\sqrt{a}$ (because it really means "a root of $x^2-a$" and it has two roots), then everything you wrote is "almost right".

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+1 for $\pm\sqrt{a}$, -1 for the reasoning at the beginning. The wrong are all but the 2nd equality. –  tohecz Dec 10 '13 at 7:02
    
@tohecz You are right. Corrected. –  Jean-Claude Arbaut Dec 10 '13 at 7:10

It is wrong because the calculation rule of square roots only works for real non negative roots.

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Well, in general, you should, when dealing with non-positive numbers, treat $\sqrt{x}$ as a set of all numbers such that $y^2=x$. Then you get that $\sqrt{1}=\{+1,-1\}$ and $\sqrt{-1}=\{+\mathrm i,-\mathrm i\}$, and then you get $$ \frac{1}{\sqrt{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \sqrt{\frac{1}{-1}} = \{+\mathrm i,-\mathrm i\}=\{\mathrm i,1/\mathrm i\},$$ and everything seems to be ok now. Of course, this brings a lot vagueness into the equals sign, similar vagueness as one has with Langrange symbols like $f(x)=\mathcal O(x)$.

As pointed in the comments, you can't write $\mathrm i=\sqrt{-1}$ just because $\mathrm i^2=-1$. Actually, $i$ is defined as such a complex number that $\mathrm i^2+1=0$, but nothing more. There are two such numbers, since the polynomial is quadratic irreducible over $\mathbb Q$.

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the inverse of $i$ is $-i$ due to the formula of

$$\frac{a+bi}{c+di}=\frac{(a+bi)(c-di)}{c^2+d^2}$$

fill in $a=d=1$ and $b=c=0$ and you'll see

$$\frac{1+0i}{0+1i}=\frac{(1+0i)(0-1i)}{0^2+1^2}=\frac{-1i}{1}=-i$$

this provides the hint that the issue is in the last square root you take, namely that you need to take the negative square root

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Square roots are not that simple with complex numbers.

You'd better use exponential form: $$i = e^{\frac{\pi}{2}i}$$ $$\frac{1}{i} = e^{-\frac{\pi}{2}i} = e^{(2\pi-\frac{\pi}{2})i} = e^{\frac{3\pi}{2}i} = -i$$

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The inverse of i is -i $$1/i = i/(i*i) = i/-1 = -i$$
so you take the negative result from \sqrt

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