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I'm not sure if I recall this correctly, but I thought there was a reason why you shouldn't write $i=\sqrt{-1}$. And if this is not true, then I wonder: Why would you define $i$ as $i^2=-1$, why wouldn't you define it as $i=\sqrt{-1}$.

I was thinking that the reason to not write $i=\sqrt{-1}$, because otherwise you could argue that $$i^2=\sqrt{-1}\sqrt{-1}=\sqrt{-1\cdot -1}=\sqrt{1}=1$$

Kasper
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6 Answers6

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None of what you wrote is the definition of the imaginary unit. The proper definition is as follows:

A complex number $z \in \mathbb C$ is an ordered pair of real numbers $(a,b) \in \mathbb R^2$ such that the following rules for addition and multiplication hold: For any $z = (a,b)$ and $w = (c,d)$ in $\mathbb C$, $$\begin{align*} w+z &= (a,b) + (c,d) = (a+c, b+d), \\ w \cdot z &= (a,b) \cdot (c,d) = (ac - bd, ad + bc). \end{align*}$$ From this definition, the imaginary unit $i$ is defined as the ordered pair $i = (0,1)$. It is then easy to show that $i^2 = -1$ using the above rules.

Masacroso
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heropup
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  • I wish my book defined it in such a clear way. Thanks for your explanation. Do you know a good book about complex analysis ? – Kasper Mar 12 '14 at 00:13
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    I'd be leery of saying 'the' proper definition when there are at least a half-dozen distinct, equivalent ways of defining the complex numbers. They all come down to this algebraically, of course, but there are subtly different motivations behind them. – Steven Stadnicki Mar 12 '14 at 00:29
  • Another good definition is $\Bbb R[x]/(x^2+1)$, where $i$ stands for the coset $x+(x^2+1)$ (and reals $r\in\Bbb R$ are abbreviations for $r+(x^2+1)$ in $\Bbb C$). – anon Mar 12 '14 at 00:32
  • Even with this construction, we could as readily define $i=(0,-1).$ – Cameron Buie Mar 14 '14 at 16:56
  • @CameronBuie Yes, you could do that but that is not how $i$ is defined. The point of a definition is that it is unambiguously stated without need for proof. We specify the structure such that there is no confusion of its meaning. The problem with saying $i = \sqrt{-1}$ is that it is not clear what we mean by this: which root do we mean? both $i^2 = (-i)^2 = -1$. As long as we axiomatically construct the complex numbers from the reals, there is no such confusion. – heropup Mar 14 '14 at 18:11
  • Defining $i=(0,-1)$ is completely unambiguous, and proceeding in this fashion will give precisely the same structure as that given by your definition. It is an atypical definition for $i,$ but it is not problematic.There are multiple ways to construct the complex numbers axiomatically from the reals, and in most, there are two options for which element we call "$i$." – Cameron Buie Mar 15 '14 at 00:18
  • For example, in the construction described by @seaturtles, we could instead let $i=-x+(x^2+1).$ – Cameron Buie Mar 15 '14 at 00:57
  • For the last time: yes you can define it that way but that is not the definition that is typically used. My only interest is in contrasting the "definition" typically given in high school treatments (the square root one) against the algebraic one that is given in analysis courses. It is as if you are trying to create a disagreement where there is none. – heropup Mar 15 '14 at 01:00
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You can write $i=\sqrt{-1}$ as long as you understand that all this means is that $i^2=-1$.

Masacroso
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lhf
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    Minor problem is that $x^2 = -1$ has two zeros, $x = \pm i$. – vonbrand Mar 12 '14 at 00:12
  • @vonbrand But $x^2=-1$ having two zeros is not a problem for writing $i^2=-1$ :-) – anon Mar 12 '14 at 00:16
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    Which one is $i$? – vonbrand Mar 12 '14 at 00:23
  • @vonbrand $i$ is $i$ and $-i$ is $-i$. Do you think the equation $i^2=-1$ is incorrect? (Note, this answer does not use the word "definition" or "define" anywhere; presumably $\Bbb C$ is assumed a priori.) – anon Mar 12 '14 at 00:30
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    @seaturtles, it certainly is correct. But it has two solutions, so it serves to describe a particularity of $i$, not to define it. – vonbrand Mar 12 '14 at 00:44
  • Yes. ${}{}{}{}$ – anon Mar 12 '14 at 00:45
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    @vonbrand It doesn't matter which root, just like in $,\Bbb C \cong \Bbb R[i]\cong \Bbb R[x]/(x^2+1),$ where $, i, :=, x + (x^2+1) ,=, x\ ({\rm mod}\ x^2!+!1).\ \ $ – Bill Dubuque Mar 12 '14 at 19:53
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Before you can say $i=\sqrt{−1}$ you need to make sure that there is some context in which $-1$ has an existent square root. For example, when speaking in context of real numbers, $\sqrt{-1}$ is not defined. Rather, defining $i^2=-1$ ensures that you don't let $i"="-i$, as $i^2=(-i)^2=-1$. See more at the accepted answer's "$\mathbf{Edit}$" of Is the square root of a negative number defined?.

Community
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While with nonnegative real numbers $a,$ we have the convention that $\sqrt{a}$ is the nonnegative number $b$ such that $b^2=a,$ there is no such convention for what the principal square root of other numbers "should" be. In particular, for any given non-zero complex $w$, there are two complex solutions to $z^2=w$--that is, there are two square roots of $w$--but only when $w$ is positive do we have a canonical choice for the principal square root of $w,$ $\sqrt{w}.$

Cameron Buie
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  • Complex powers are defined by $a^b:=\exp(b\log a)$ using a choice of branch for $\log$. Using the principal branch for $\log$, if $z=re^{i\theta}$ with $\theta\in(-\pi,\pi]$ then $\sqrt{z}:=\sqrt{r}e^{i\theta/2}$ is the best choice of thing to call the principal square root of the complex number $z$ (I believe this is used to some extent as a standard; I didn't just make it up). – anon Mar 12 '14 at 00:14
  • That is certainly the common choice (and what I typically use), but that branch cut is arbitrary, and there's no "obvious" reason why we shouldn't take $\theta\in[-\pi,\pi),$ for example, which of course leads to a different principal square root function. – Cameron Buie Mar 15 '14 at 18:09
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If you have done some abstract algebra, another definition of $\mathbf{C}$ is that it is the quotient ring $\mathbf{R}[X]/(X^2+1)$. The imaginary unit is then the coset of $X$ in this ring, and we then have $\overline{X}^2 + 1 = \overline{X^2+1} = \overline{0}$, so $\overline{X}^2 = -1$. (Relate this definition with the one given by heropup.)

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fkraiem
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Complex numbers are not an ordered ring, so in other words giving them an order doesn't make sense. In fact neither of those definitions defines the element uniquely. Since if $i^2=-1$ then $(-i)^2=(-1(i))^2=(-1)i(-1)i=(-1)(-1)i^2=1(-1)=-1$

Asinomás
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