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let $i$ be a complex number (Unit imaginary part) , Really I don't have a convincing method for showing to my student that $ \sqrt{i^2} \neq(\sqrt{i})^2 $ , because he know that for $x$ positive real number we have :$$ \sqrt{x^2} =(\sqrt{x})^2 $$, Any way ?

zeraoulia rafik
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    Fractional powers of negative numbers aren't uniquely defined; cf. this question – J. W. Tanner Feb 17 '20 at 22:01
  • Maybe matheducators.se is the place for this question. – J.G. Feb 17 '20 at 22:11
  • It's unclear what you're looking for here. With the branch cut of the square root in the lower half of the complex plane, the identity you want to disprove is in fact true, and I would consider this to be the "standard" convention. I think your efforts are misguided. – Yly Feb 17 '20 at 22:19

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With the usual main branch of the complex square root,

$$\sqrt{i^2}=\sqrt{-1}=i$$

and

$$(\sqrt i)^2=\left(\frac{1+i}{\sqrt2}\right)^2=\frac{1+2i+i^2}2=i.$$

So I don't think you can show them.

If your claim is related to the fact that a number has two square roots, then you can write $$\pm\sqrt{i^2}=\pm\sqrt{-1}=\pm i$$

while

$$(\pm\sqrt i)^2=\left(\pm\frac{1+i}{\sqrt2}\right)^2=i,$$ which brings no visible contradiction.

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Indeed, when you work in the complex domain, the square root does not have a nice definition in the whole plane as a single-valued function. For every complex number z different from zero the equation w^2 = z has two different solutions. In order to have a single-valued function, ¡you neeed to choose one! And you cannot do that in a continuous way (less to say holomorphic way) in the whole plane.

Usually one makes a cut along a line, for example the negative real axis (but this would leave the square root of -1 undefined)

The origin z=0 is a branch point of the square root (https://en.wikipedia.org/wiki/Branch_point).

Pablo De Napoli
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The identity $(\sqrt x)^2=x$ holds for all complex numbers $x$; it's the equation $\sqrt{x^2}=x$ that sometimes holds and sometimes doesn't. The set where it doesn't hold depends on the convention that one chooses to use for defining the square root symbol as a single-valued function. In all standard conventions, however, we have $\sqrt{-1}=i$ (rather than $-i$), and, consequently, we have

$$\sqrt{i^2}=\sqrt{-1}=i$$

and so in fact we do have $\sqrt{i^2}=(\sqrt i)^2$ (provided we're using a standard convention).

The question to ponder with your student is precisely why does $(\sqrt x)^2=x$ hold for all complex numbers $x$, but $\sqrt{x^2}=x$ cannot always hold?

Barry Cipra
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  • I don't know what you mean by standard conventions. Could you please provide a reference? – Pablo De Napoli Feb 20 '20 at 21:17
  • @PabloDeNapoli, the two most common conventions are easiest to state in polar coordinates: $\sqrt{re^{i\theta}}=\sqrt re^{i\theta/2}$ with $0\le\theta\lt2\pi$ for one convention and $-\pi/2\lt\theta\le\pi/2$ for the other (and $r\ge0$, of course). You sometimes see this described in terms of where the "branch cut" is made. It's basically a question of whether you want your square roots to have positve real part or positive imaginary part. – Barry Cipra Feb 20 '20 at 21:38