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I have a small point to be clarified.We all know $ i^2 = -1 $ when we define complex numbers, and we usually take the positive square root of $-1$ as the value of "$i$" , i.e, $i = (-1)^{1/2} $.

I guess it's just a convention that has been accepted in maths and the value $i = -[(-1)^{1/2}] $ is neglected as I have never seen this value of "$i$" being used. What I wanted to know is, if we will use $i = -[(-1)^{1/2}] $ instead of $ (-1)^{1/2} $, would we be doing anything wrong? My guess is there is nothing wrong in it as far as the fundamentals of maths goes. Just wanted to clarify it with you guys. Thanks.

Jonas Meyer
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under root
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... and we usually take the positive square root of −1 as the value of "$i$".

There's no such thing as a positive or negative complex number. By treating $\mathbb{R}$ as a subset of $\mathbb{C}$ you can call some complex numbers positive or negative, but only those which actually are real numbers. There's no way to define a total order on $\mathbb{C}$ which would behave like the usual order on $\mathbb{R}$ does.

What happens is that one simply picks any root of $-1$, i.e. any solution of $x^2 + 1 = 0$, and calls it $i$. There's then always a second solution, and that solution is $-i$. You can't even say which solution you picked for $i$ - the two solutions of $x^2 + 1=0$ are algebraically indistinguishable, i.e. you cannot tell them apart with algebraic means.

Imagine a friend hands you a bag containing two balls, of equal size and material. You can pick one arbitrarily and call it "ball 1", and take a pen and mark it with a big "1". The other is then "balls 2", and gets marked with a big "2". Now, imagine you had picked the other ball. Would you end up in a different situation? No! You'd still have two balls, one marked "1" and one marked "2", and indistinguishable otherwise. Now, after you've marked the balls, they are of course different. You can now for example put both back into the bag, let your friend pick one, and ask "Which ball have you picked?". But that only works after you marked them!

fgp
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  • ,thanks for your feedback. So,one can define i = $ -(-1)^{1/2} $ as a solution to $ i^2 $ = -1 and can take -i = $ (-1)^{1/2} $ ... right? there is nothing wrong in it... – under root May 20 '13 at 12:58
  • i mean only if one wants to...though its better to follow what has been followed since years... – under root May 20 '13 at 12:59
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    @under-root Well... no, not really. I suggest you re-read my last paragraph. The whole point is that $\sqrt{-1}$ isn't a uniquely defined thing. It's a charaterization of a thing, as a solution of $x^2 = -1$. But there are two things which fullfill this equation, and both are completely indistinguishable. $i$ is an arbitrary one. $-i$ is then the other one. You can't say which one $i$ is, same as you can't say which ball you marked with "1". – fgp May 20 '13 at 13:29
  • @under-root You can, of course, first pick an arbitrary one and call it $\sqrt{-1}$. Then you can set $i = -\sqrt{-1}$, i.e. call the other one $i$. – fgp May 20 '13 at 13:31
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    @under-root The point is, $\sqrt{-1}$ isn't any particular solution. It's an arbitrary one. $i$ is an arbitrary one too. This leaves you with the choice of whether $\sqrt{-1}=i$ or $\sqrt{-1}=-i$. Usually one picks the former, but you can of course pick the latter if you're so inclined... – fgp May 20 '13 at 13:34
  • ok... thanks a lot for your valuable suggestions.Got it now :)) – under root May 20 '13 at 18:36
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The square roots with the properties we know are used only for Positive Real numbers.

We say that $i$ is the square root of $-1$ but this is a convention. You cannot perform operations with the usual properties of radicals if you are dealing with complex numbers, rather than positive reals.

It is a fact that, $-1$ has two complex square roots. We just define one of them to be $i$.

You have to regard the expression $i=\sqrt {-1}$ just as a symbol, and not do operations.

Counterexample: $$\dfrac{-1}{1}=\dfrac{1}{-1}\Rightarrow$$ $$\sqrt{\dfrac{-1}{1}}=\sqrt{\dfrac{1}{-1}}\Rightarrow $$ $$\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}\Rightarrow$$ $$\dfrac{i}{1}=\dfrac{1}{i}\Rightarrow$$ $$i^2=1\text { ,a contradiction}$$

Dimitris
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  • Could you please be more specific as to what operations you are referring to that we cannot do? – Jonas Meyer May 20 '13 at 12:11
  • @JonasMeyer does the example I added supplement this? – Dimitris May 20 '13 at 12:20
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    I agree with the part about having to define $i$ to be one of the square roots of $-1$ and sticking with the choice. However, I don't see how the part about being careful with operations is particularly relevant to the question. You say "counterexample," but it is not a counterexample to anything in the question as far as I can see, but rather to the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ which would hold for the usual square root function if $a\geq 0$ and $b>0$. – Jonas Meyer May 20 '13 at 12:21
  • I want to make clear the fact that $i=\sqrt{-1}$ cannot be used that way in operations. So, does it make sense to take the other root $i = -[(-1)^{1/2}]$? – Dimitris May 20 '13 at 12:24
  • If we have a definition of $\sqrt{-1}$ in the first place (as a particular number whose square is $-1$), but have not yet designated our definition of $i$, then yes, it would make perfect sense to define $i=-\sqrt{-1}$ (the other number whose square is $-1$), and then one could carry out all of the theory of complex numbers with this choice. So there is a bit more to the question in that sense. – Jonas Meyer May 20 '13 at 12:26
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    Yes, I completely agree with that. So, is there a real difference between $i = -[(-1)^{1/2}] $, and $i = [(-1)^{1/2}] $? We sometimes think in the sense of Real numbers and operations. That's what I wanted to point out. – Dimitris May 20 '13 at 12:28
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You are correct that $\sqrt{-1} = -i$, but you are not correct when you say that this fact is neglected--it's used all the time. E.g. the quadratic formula:

$$x^2+x+1=0$$

$$x=\frac{1}{2(1)}+ \frac{\sqrt{1^2-4(1)(1)}}{2(1)}$$

$$x=\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$$

The plus-or-minus is there because $\sqrt{-3}$ has two roots: $i\sqrt{3}$ and $-i\sqrt{3}$.

rurouniwallace
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    If $\sqrt{-1}$ and $i$ are used to mean the same thing, as they are in many contexts including the beginning of the question above, then the equation $\sqrt{-1}=-i$ is incorrect. You seem to be using the square root symbol to denote a 2-valued function, but this does not clear up the confusion in the question as far as I can see. – Jonas Meyer May 20 '13 at 12:15
  • Strictly speaking, the square root operation (that is, $\sqrt{x}$) always takes the "positive" value. This is why the $\pm$ is usually included in the quadratic formula. – Glen O May 20 '13 at 12:17
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    Unfortunately $i$ is not positive in any senseful meaning. – Andrea Mori May 20 '13 at 12:21
  • @GlenO Ok, I see what the confusion is now, but as I understand it $y=\sqrt{x}$ is anything that satisfies $y^2=x$, unless it's explicitly stated that $\sqrt{x}$ is intended to be used as a function. – rurouniwallace May 20 '13 at 12:24
  • @ZettaSuro: The "square root" is anything that satisfies that equation, but $\sqrt{x}$ is taken to be the positive value. That is, $\sqrt{x^2} = |x|$. – Glen O May 20 '13 at 12:33
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    @Glen0: What you just wrote only makes sense for square roots of nonnegative numbers. In particular, note that $|x|$ is not a square root of $x^2$ unless $x$ is real. – Jonas Meyer May 20 '13 at 12:35
  • @GlenO I respectfully disagree. – rurouniwallace May 20 '13 at 12:43
  • @JonasMeyer: True. I was referring to the real case, and should have said as much (oversight on my part). But the point I was making remains - that the square-root symbol is taken to define the "positive" square root function. – Glen O May 20 '13 at 12:49
  • @ZettaSuro: This isn't an opinion, it's a statement of fact. The square root operation (that is, $\sqrt{x}$) is taken to be positive specifically because otherwise it would be necessary, for instance, to go $\sqrt{|\sqrt{x}|}$ in order to get real fourth roots of positive numbers, and one would never see the $\pm$ written in the quadratic equation without that restriction (I've never seen it left off). – Glen O May 20 '13 at 12:51
  • You incorrectly omitted the $\pm$ sign in the quadratic formula. (The quadratic formula always uses a $\pm$ sign because $\sqrt{}$ is a functional operator and can only return one value). So the convention $\sqrt{-1}=-i$ is not being used here. The $\pm$ symbol is simply being carried down to your last line, leading to the conjugates that you wrote. So you are incorrect when you say that \sqrt{} (without the $\pm$ sign) can return two values. – Radial Arm Saw Jul 12 '20 at 22:16
  • @Ataraxia $\sqrt{}$ is defined to be a function so it must always be a function and only returns the principle value. – Radial Arm Saw Jul 12 '20 at 22:22