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this is something that came up when working with one of my students today and it has been bothering me since. It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.

Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?

With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?

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    $\begingroup$ It all depends on how you define $\sqrt{x}$. $\endgroup$ – Thomas Andrews Jan 9 '15 at 0:26
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    $\begingroup$ And it is not true that positive number square roots "can return positive and negative." That depends on how you define $\sqrt{x}$ there, too. $\endgroup$ – Thomas Andrews Jan 9 '15 at 0:27
  • $\begingroup$ funny I thought that negative square roots was impossible $\endgroup$ – Irrational Person Jan 9 '15 at 0:31
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    $\begingroup$ When I introduce the complex numbers, I never use $\sqrt{-1}=i$, but say that $i^2=-1$, which is different. There are two complex numbers whose square is $-1$. $\endgroup$ – egreg Jan 9 '15 at 0:33
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    $\begingroup$ People who say $\sqrt{-1} = i$ are being sloppy, either intentionally or unintentionally. It is false that $\sqrt{-1} = i$. It is true that $i^2 = -1$, but it is also true that $(-i)^2 = -1$. $\endgroup$ – MJD Jan 9 '15 at 0:37
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The square root function doesn't return two values for positive numbers, or it wouldn't be a function.

It's a fact that, if $x$ is a positive real number, there are two real numbers whose square is $x$. The positive one is denoted by $\sqrt{x}$, so the negative one is $-\sqrt{x}$.

In this way the function has the pleasant property that, for $x,y>0$, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.

In the complex numbers, for every nonzero $x\in\mathbb{C}$ there are two complex numbers whose square is $x$. However, it's not possible to define a square root function with the property above, that is, $\sqrt{xy\mathstrut}=\sqrt{x\mathstrut}\sqrt{y\mathstrut}$.

Maybe you'd want to stretch the notion of function, to allow multiple values; but then, how many values should you assign to the expression $$ \sqrt{2}+\sqrt{3}+\sqrt{5} $$ and similar ones? You couldn't make the more obvious simplifications: from $$ x+\sqrt{2}=\sqrt{2} $$ you'd get three values for $x$.

When I introduce the complex numbers, I never use $\sqrt{-1}$, but rather I say that $i^2=-1$, which is a quite different statement, exactly because it's impossible to define a square root function that has sensible algebraic properties.

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  • $\begingroup$ Actually, it is true that $\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}$ where equality is in the sense of set equivalence. Of course, the branch of the square root operators might need to be chosen differently from one another. $\endgroup$ – Mark Viola Aug 14 '19 at 20:56
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If $x$ is a non-negative real number, there's an unambiguous interpretation for the expression $\sqrt{x}$, namely, the non-negative square root of $x$. (The $\pm$ signs aren't "part of the square root function", which is why they have to be included explicitly when "solving an equation by taking square roots", e.g., passing from $x^{2} = 1$ to $x = \pm 1$.)

When one tries to extend the square root function to the complex numbers, there are tricky domain issues. Rather than write "$\sqrt{-1} = \pm i$" (which can't be true if the radical symbol denotes a function), it's safer to stick with $(\pm i)^{2} = -1$ (which is unambiguously true) until one is invested in understanding the fine points of functions of a complex variable.

In case you or your student are curious: Each non-zero complex number has two square roots, but there is no continuous choice of square root on the complex plane.

To get a continuous "branch of square root" it's necessary to remove enough of the plane that "the domain of the square root doesn't encircle the origin". The customary choice is to remove the non-positive reals. (Ironically, this explicitly excludes $-1$ from the domain of the square root.)

A common alternative choice is to remove the non-positive imaginary axis. In this event, $\sqrt{-1} = i$ for the continuous branch of square root that satisfies $\sqrt{1} = 1$.

The take-away points are:

  • If $\sqrt{\ }$ denotes a function, then it must be single-valued (no $\pm$).

  • When allowing complex numbers (other than non-negative reals) under a radical sign, You Really Need To Be Careful.

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Well, if you are considering that $y=\sqrt{x}$ is the relation $y^2=x$, then, yes, $\pm i$ are both solutions to $\sqrt{-1}$. However, this is not usually how square roots are defined. Typically we say: $$\sqrt{1}=1$$ Not plus or minus $1$ - just $1$. This means that $\sqrt{x}$ is a "right inverse" of $x^2$ - that is, we have that $$\left(\sqrt{x}\right)^2=x$$ but not necessarily that $$\sqrt{(x^2)}=x.$$ The difference here is that, to make $\sqrt{x}$ a function, we need it to yield unique output values - so we choose a principal root. We generally define that whenever we take the square root of a positive number, we get a positive number. And the definition $\sqrt{-1}=i$ is equally innocuous, since if we defined it as $\sqrt{-1}=-i$, we could just relabel every number by its complex conjugate, and get back to our typical definition (that is to say, $\sqrt{-1}=i$ more defines $i$ than it defines $\sqrt{-1}$). So, though the equation $x^2+1$ has two solutions (as does any equation $x^2-c$ for complex $c\neq 0$), for the sake of making square roots act like a function, we must choose one of them to be the square root of $-1$, and we call this number $i$.

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    $\begingroup$ "usually" is context-dependent. In complex analysis, $\sqrt{}$ is often considered to be a multivalued function. $\endgroup$ – Robert Israel Jan 9 '15 at 0:43
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This question can be approached from one of two angles. Let's do one at a time.

  1. Can $\sqrt{-1}$ be both $i$ and $-i$?

The answer is in fact yes. People realized early on that some functions may take any of a number of values. This is true already in the real number system with the square root, but usually well-behaved functions like the logarithm also become "multi-valued" when extended to take as input complex numbers. The solution is to change the domain of the function, meaning that the square root takes input from "two copies of the complex numbers sewn together", something called a Riemann surface.

  1. Are $\sqrt{-1}=i$ and $\sqrt{-1}=-i$ in some sense "the same"?

Again, the answer is yes. As Meelo remarked in his answer, both $i$ and $-i$ are solutions to the polynomial equation $x^2+1=0$. However, they are equivalent solutions, in the sense that interchanging them generates a symmetry of the complex plane. This symmetry is described by flipping the plane about the real axis, also known as complex conjugation. This is the starting point of Galois theory.

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There are many abstract ways to construct the complex numbers, and however you do it, there is a complex number which squares to $-1$, and to which you happen to give a name, often $i$. Once you have named $i$, it becomes obvious that $-i$ also has square $-1$, and that all complex numbers have the form $a +bi$ for unique real $a$ and $b.$ It really does not matter which $i$ you choose in the first place, you could equally well have chosen $-i$ instead, and this is reflected by the fact that the mapping which sends $a + bi$ to $a-bi$ for $a,b \in \mathbb{R}$ preserves all the additive and multiplicative structure of the complex field (ie is an automorphism of the complex field fixing every element of $\mathbb{R}$). The point is that there is no mystical significance to the symbol $i$. If you make the other choice and give $-i$ a new symbol $j$, you can work equally well with expressions of the form $a + bj$ for $a,b \in \mathbb{R}$, and never know the difference.

When you start to represent complex numbers geometrically on the Argand diagram, you do tend to make a choice, but that is a choice of using clockwise or counterclockwise direction for rotations. It is usual to represent the complex number $\cos \theta + i\sin \theta$ by the point $(\cos \theta, \sin \theta)$ in $\mathbb{R}^{2}.$ But if we consistently used $(\cos \theta, -\sin \theta)$ instead we would have no difficulty.

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To be honest, I would refrain from saying that $$\sqrt{-1} = i$$ It's not even like the real numbers, where one can agree that the square root function takes on the positive values, because what is positive, in the complex plane? Even if you did establish the identity as convention, how would you extend the convention when dealing with: $$\sqrt[\Large 3]{1}$$ for example?

If you need to define the imaginary unit, the best choice is to say that $i$ is such that $$i^2 = -1 $$

There are even "better" choices, but it doesn't look like you'll be needing them.

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In fact, there are two complex square roots of -1. i and -i. So in fact the answer to your question that it is equal to +i or -i..

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