Why is \sqrt{-x^{2}}=|x|i So, i=\sqrt{-1}

Step 1Originally, 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.Step 2If 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\ge 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).Step 3Once 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\ge 0$ then $|x|=x$ and this is what you get, and if $x\le 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)}{=}^{\text{not really}}-\sqrt{-1}\sqrt{-1}=i\times 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 $\varphi$ (in radians) such that$z=\rho (\cos \theta +i\sin \theta )$$w=\sigma (\cos \varphi +i\sin \varphi )$If you do that, then it is easy to check that$zw=(\rho \sigma )(\cos (\theta +\varphi )+i\sin (\theta +\varphi ))$;$\sqrt{z}=(\sqrt{\rho }(\cos \frac{\theta }{2}+i\sin \frac{\theta }{2})),(\sqrt{\rho }(\cos (\frac{\theta }{2}+\pi )+i\sin (\frac{\theta }{2}+\pi ))):$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).Step 4So, if$x=\rho (\cos \theta +i\sin \theta )$, then$-x^2={\rho }^2(\cos (2\theta +\pi )+i\sin (2\theta +\pi ))$(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${\sqrt{\rho }}^2(\cos (\frac{2\theta +\pi }{2})+i\sin (\frac{2\theta +\pi }{2}))=\rho (\cos (\theta +\frac{\pi }{2})+i\sin (\theta +\frac{\pi }{2}))$

Step 1 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 $-a<0$ $\sqrt{-a}=\sqrt{-1\times a}=i\sqrt{a}$ Note that the first rule required that the two numbers were nonnegative, so $\sqrt{-4}\sqrt{-9}=2i\times 3i=2\times 3\times i^2=6\times -1=-6$ is not equal to $\sqrt{-4\times -9}=\sqrt{36}=6$ 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, $\sqrt{-x^2}=i\sqrt{x^2}=i|x|$

Step 1 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. Step 2 This is essentially a matter of definitions. If x is a complex number, and $x\ne 0$, we can write $x=r{e}^{i\theta }$ 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 $\sqrt{r}{e}^{\frac{i\theta }{2}}$ 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 $i=\sqrt{-1}=e^{\frac{i\pi }{2}}$ Then $\sqrt{-x^2}=\sqrt{r^2{e}^{i(2\theta +\pi )}}$. Now, if $0\le \theta <\frac{\pi }{2}$ (i.e., if x is in the first quadrant), then $2\theta +\pi$ is the principal argument of $-x^2$, and $\sqrt{-x^2}=r{e}^{i(\theta +\frac{\pi }{2})}=ix$ 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 $\frac{\pi }{2}\le \theta <2\pi$, 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: $\sqrt{-x^2}=r{e}^{i(\theta -\frac{\pi }{2})}=ix$ In the second, we get another instance where the identity holds: $\sqrt{-x^2}=r{e}^{i(\theta -\pi -\frac{\pi }{2})}=x(-1)(-i)=ix$ Finally, note that we are in the first case above iff $0\le 2\theta -\pi <2\pi$, i.e., iff $\frac{\pi }{2}\theta <\frac{3\pi }{2}$ (i.e., if $x=a+bi$ where a,b are real, and either $a=0\mathrm{\&}b>0,$ or $a<0$), while we are in the second case iff $\frac{3\pi }{2}\le \theta <2\pi$.