# Why is $\sqrt{-x^2}=|x| i$

So $i = \sqrt{-1}$

But why is $\sqrt{-x^2}=|x| i$

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This should be $\sqrt{-x^2}=xi$. – Vitaly Lorman Jan 10 '11 at 17:50
$\sqrt{-x}= \imath \cdot \sqrt{x}$ – Bryan Yocks Jan 10 '11 at 17:55
Vitaly: You are right. I've fixed it. – paldepind Jan 10 '11 at 18:33
the title should be appropriately edited as well. – Sean Tilson Jan 10 '11 at 18:46

Edit: Originally, we had "$\sqrt{-x} = i\sqrt{x}$" (or rather, the incorrect $\sqrt{-x}=ix$), and I answered using that; I edited this to account for the correction made to the question.

It depends on what kind of animal $x$ is and what you mean by $\sqrt{x}$. (In other words: don't believe everything Google tells you). For example, say $x=i$; then $|x|=1$, $x^2=-1$, so $\sqrt{-x^2}$ is $\sqrt{1}$, whereas $|x|i=i$. And, no matter how you interpret the square root, it is certainly not true that $\sqrt{1}$ is equal to $i$.

So, what is going on here?

If $x$ is a real number, then $x^2$ is nonnegative, so that $\sqrt{-x^2}$ is asking you to take the square root of a negative real number.

But when dealing with complex numbers, we really do need to keep in mind that the square root function has two "branches"; you may remember that in the real number case, one starts with the function $y=x^2$, which is not one-to-one (does not pass the horizontal line test); in order to get an inverse, we restrict the domain to $x\geq 0$ so that the resulting restricted function is one-to-one, and $\sqrt{x}$ is defined to be the inverse of this function. In particular, $\sqrt{x}$ only accepts nonnegative numbers as inputs, and only gives nonnegative numbers as outputs. Because $\sqrt{x}$ is always nonnegative, it has some nice properties (chief among them: it is a function; also, $\sqrt{ab}=\sqrt{a}\sqrt{b}$ if $a$ and $b$ are both nonnegative real numbers).

Once you go to complex numbers things get more complicated, because there no longer is, in general, a good way of selecting which of the two solutions to $y^2=a$ you want to call "the" square root of $a$. This means that we really do need to keep track of both solutions, otherwise things get really messy and you end up with nonsense (as you will see below). So when dealing with square roots of complex numbers we really have what is sometimes called a "multi-valued function", or a function with "two branches": there are really two possible values for $\sqrt{a}$ when $a$ is a complex number. This includes real numbers when you allow negative ones as inputs.

If $x$ is a real number, then $x^2$ is nonnegative. Then the two the two complex square roots of $-x^2$ are $i|x|$ and $-i|x|$; in fact, you can just take $ix$ and $-ix$, because if $x\geq 0$ then $|x|=x$ and this is what you get, and if $x\leq 0$ then $|x|=-x$, so $i|x|=-ix$ and $-i|x|=ix$. To verify these are the two values of the (multi-valued) square root, square both and you'll see that that they both give $-x^2$. So one branch gives you the value $ix$ (or $i|x|$), and the other branch gives you the value $-ix$ (or $-i|x|$).

Note well: You have to be careful with the complex square root function: in particular, it is no longer true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ always holds: for instance, putting $a=b=-1$, you would get silly things like: $$1 = \sqrt{1} = \sqrt{(-1)(-1)} \stackrel{\text{not really}}{=}\sqrt{-1}\sqrt{-1} = i\cdot i = -1.$$

If $x$ is a nonreal complex number, though, you have to be more careful. If you write complex numbers in polar form, then multiplication and square roots are easy. If $z$ and $w$ are complex numbers, you can always find nonnegative real numbers $\rho$ and $\sigma$, and angles $\theta$ and $\phi$ (in radians) such that \begin{align*} z &= \rho\left(\cos\theta + i\sin\theta\right)\\ w &= \sigma\left(\cos\phi + i\sin\phi\right). \end{align*} If you do that, then it is easy to check that \begin{align*} zw &= (\rho\sigma)\left(\cos(\theta+\phi) + i\sin(\theta+\phi)\right);\\ \sqrt{z} &= \left\{\begin{array}{l} \sqrt{\rho}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right),\\ \sqrt{\rho}\left(\cos\left(\frac{\theta}{2}+\pi\right) + i\sin\left(\frac{\theta}{2}+\pi\right)\right). \end{array}\right. \end{align*} Above, $\sqrt{\rho}$ is the nonnegative real number whose square is equal to $\rho$ (remember that $\rho$ is a real number greater than or equal to $0$).

So, if $x=\rho(\cos\theta+i\sin\theta)$, then $$-x^2 = \rho^2\left(\cos(2\theta+\pi) + i\sin(2\theta+\pi)\right)$$ (I used the fact that $-1 = 1(\cos\pi + i\sin\pi)$ and the formulas above). So the two values of the square root are \begin{align*} \sqrt{\rho^2}\left(\cos\left(\frac{2\theta+\pi}{2}\right) + i\sin\left(\frac{2\theta+\pi}{2}\right)\right) &= \rho\left(\cos\left(\theta+\frac{\pi}{2}\right) + i\sin\left(\theta+\frac{\pi}{2}\right)\right),\\ \sqrt{\rho^2}\left(\cos\left(\frac{2\theta+\pi}{2}+\pi\right) + i\sin\left(\frac{2\theta+\pi}{2}+\pi\right)\right) &= \rho\left(\cos\left(\theta+\frac{3\pi}{2}\right) + i\sin\left(\theta+\frac{3\pi}{2}\right)\right). \end{align*} On the other hand, multiplying $x$ by both $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$ and $-i=\cos\frac{3\pi}{2}+ i\sin\frac{3\pi}{2}$ gives the two values: $$\rho\left(\cos\left(\theta+\frac{\pi}{2}\right)+i\sin\left(\theta+\frac{\pi}{2}\right)\right)\text{ and }\rho\left(\cos\left(\theta+\frac{3\pi}{2}\right)+i\sin\left(\theta+\frac{3\pi}{2}\right)\right)$$ so that these two are still the correct answers.

How does this relate to the error I noted above? If $x=i$, then the two complex branches of the square root give that the two values of $\sqrt{-x^2}$ are $1$ and $-1$. If we look at the two solutions for the general case, which are $ix$ and $-ix$ (notice that I cannot use the absolute value bars if we don't assume $x$ is real to begin with), then we get $i(i) = -1$ and $-i(i) = 1$, the same tow answers (though in different order). So keeping track of both branches is paramount in the complex case.

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For taking square roots of complex numbers it is quicker to find one, say $w$, the other will be $-w$. The fundamental theorem of algebra implies there are no more. – Jp McCarthy Nov 23 '11 at 1:18

First, if $a$ and $b$ are nonnegative real numbers ($a,b\in\mathbb{R}$, $a,b\ge 0$), $\sqrt{a}\sqrt{b}=\sqrt{ab}$. To deal with a square root of a negative real number, say $-a<0$, $\sqrt{-a}=\sqrt{-1\cdot a}=i\sqrt{a}$. Note that the first rule required that the two numbers were nonnegative, so $\sqrt{-4}\sqrt{-9}=2i\cdot 3i=$ $2\cdot3\cdot i^2=$ $6\cdot-1=-6$ is not equal to $\sqrt{-4\cdot-9}=\sqrt{36}=6$.

Second, note that for $x\in\mathbb{R}$, $\sqrt{x^2}=|x|$. For example, $\sqrt{(-3)^2}=\sqrt{9}=3$.

Now, for $x\in\mathbb{R}$, $x^2\in\mathbb{R}$ and $x^2\ge 0$. So, $\sqrt{-x^2}=i\sqrt{x^2}=i|x|$.

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Alternatively, note that $(xi)^2 = x^2i^2 = -x^2$, therefore $ix$ is a square root of $-x^2$. – Myself Jan 10 '11 at 18:19
@Myself: a square root is a start, but there should also be some justification as to why it's the principal square root (the one indicated by the radical symbol). – Isaac Jan 10 '11 at 18:30
You're right! But as stated in many answers: that's a convention and nothing more. I've seen both signs in $\sqrt{-i}=\pm \frac{1-i}{2}$, depending on whether you cut along the positive or negative real axis (i.e whether you take the argument to be in $[0,2\pi)$ or $(-\pi/2,\pi/2]$). I just tend to avoid the $\sqrt{}$-sign. And if I do use it, it 's just a square root and the argument works for every choice. – Myself Jan 10 '11 at 21:23
@Myself: I've discussed the lack of a standard definition of principal root for nonreal complex numbers myself on several occasions, but there is a clear definition of principal square root of a real number: nonnegative real numbers have nonnegative real principal square roots; negative real number have principal square roots along the positive imaginary axis (defining $i=\sqrt{-1}$ as the principal root of $-1$ strongly suggests that the rest of the principal roots of negative reals should be positive real constant multiples of it—the choice is in which root is represented by $i$). – Isaac Jan 10 '11 at 22:05
Let me just say I agree with this post (goo.gl/rybue) on the matter, stating $\sqrt{-1}$ is comfortable but imprecise. Therefore I think it should be avoided in a context of precalculus or elementary algebra. – Myself Jan 11 '11 at 2:02

Edit: The original question asked why $\sqrt{-x^2}=xi$. This is false in general, as I answered below. Here I just address the new version of the question (using what I wrote below).

Note that we have, from what I wrote below, that $\sqrt{-x^2}=\pm xi$ for all $x$, with the choice of $+$ or $-$ depending on where $x$ is in the plane. if $x$ happens to be real, $x=r^{e^{i\theta}}$ where $\theta=0$ or $\pi$, and the above is just $\sqrt{-r^2}=ir=i|x|$.

If $x$ is not real, the equality fails.

This is essentially a matter of definitions. If $x$ is a complex number, and $x\ne 0$, we can write $x=re^{i\theta}$ for some positive real number $r$ and some real number $\theta$. There are many choices of $\theta$, but the principal argument is defined so $0\le\theta<2\pi$. Here, $e^{i\theta}=\cos(\theta)+i\sin(\theta)$.

We define $\sqrt x$ as $\sqrt r e^{i\theta/2}$ where $\sqrt r$ is the (real) positive square root of $r$, and $\theta$ is the prinicipal argument. Also, we define $i$ as $\sqrt{-1}$, which according to the conventions just described, gives us: $-1=e^{i\pi}$ so $i=\sqrt{-1}=e^{i\pi/2}$.

Then $\sqrt{-x^2}=\sqrt{r^2e^{i(2\theta+\pi)}}$. Now, if $0\le \theta<\pi/2$ (i.e., if $x$ is in the first quadrant), then $2\theta+\pi$ is the principal argument of $-x^2$, and $$\sqrt{-x^2}=re^{i(\theta+\pi/2)}=ix.$$

However, this identity does not always hold (Google notwithstanding) if $x$ is not in the first quadrant. For example, if $x=-1$, then $\sqrt{-x^2}=\sqrt{-1}=i$ while $ix=-i$. The reason is that if $\pi/2\le \theta<2\pi$, then $2\theta+\pi$ is not the principal argument of $-x^2$, the principal argument will be $2\theta-\pi$ or $2\theta-3\pi$, whichever lands us again in the range ${}[0,2\pi)$.

In the first case, we always get a failure of the identity: $$\sqrt{-x^2}=re^{i(\theta-\pi/2)}=-ix.$$

In the second, we get another instance where the identity holds: $$\sqrt{-x^2}=re^{i(\theta-\pi-\pi/2)}=x(-1)(-i)=ix.$$

Finally, note that we are in the first case above iff $0\le2\theta-\pi<2\pi$, i.e., iff $\pi/2\le\theta<3\pi/2$ (i.e., if $x=a+bi$ where $a,b$ are real, and either $a=0\&b>0$, or $a<0$), while we are in the second case iff $3\pi/2\le\theta<2\pi$.

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I think that this should be addressed in a much more fundamental level:

Suppose $x\in\mathbb{R}$ then $\sqrt{x^2} = |x|$.

The reason this is happening, is that we want the $\sqrt{\cdot}$ function to have certain properties, which cannot be well-defined on real numbers otherwise.

First we want it to be a function, which means that given $x$ there is exactly one $y$ such that $\sqrt{x} = y$. So we can choose either the positive or the negative root - but not both of them, so we take the positive one.

We want to have $\sqrt{x} = x^{\frac{1}{2}}$ so we want to have roots respecting the general laws of exponentiation - for example $(x^m)^n = x^{m\cdot n}$, so we want $(x^{\frac{1}{2}})^{\frac{1}{2}} = \sqrt{\sqrt{x}} = \sqrt[4]{x} = x^{\frac{1}{4}}$, but suppose we had chosen $\sqrt{x^2} = -|x|$ then while we could theoretically define $\sqrt[4]{x^2}$ we could still not define within the real numbers $\sqrt{-|x|}$ and thus the above equality wouldn't hold.

Lastly, people often think of the $\sqrt{x}$ as an inverse of the $x^2$ function, but people are often forgetting (or unaware) that inverse can only be defined when the function is injective, namely that $f(x)=f(y)$ happens only if $x=y$ (for example, $f(x)=x+1$ has this property). However within the realm of the real numbers this is not the case, as $(-2)^2 = (2^2) = 4$, and therefore the $x^2$ cannot be inversed unless we bound ourselves to a subset of the real numbers on which squaring a number is indeed injective - e.g. the non-negative real numbers.

This is just one of the caveats that one must learn to avoid when doing mathematics properly.

Now for the sake of completeness, to address your question - based on the aforementioned arguments - we have (for $x\in\mathbb{R}$):

$\sqrt{-x^2} = \sqrt{-1}\cdot\sqrt{x^2} = i|x|$

The problem defining the root (in fact more than just square root) over complex numbers is essentially the same, only that in the complex numbers you have $n$ roots for $\sqrt[n]{x}$ and you must define properly which one you're taking.

Suppose we take the least non-negative $\theta$ such that $(\sqrt{r}e^{i\theta})^2$ is $x$ (that is, we choose a root, just like we did with the real numbers by choosing the positive root) then we have:

$\sqrt{-r^2e^{2i\theta}} = \sqrt{r^2e^{2i(\theta+\frac{\pi}{2})}} = re^{i(\theta+\frac{\pi}{2})}=ire^{i\theta}$

However $|re^{i\theta}|=r$, and so the equality above is usually not true for complex numbers (however it can be restated in such way which generalizes it for the above choice of complex root).

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