How I convince my student that $ \sqrt{i^2} \neq (\sqrt{i})^2 $? 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 ?
 A: 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.
A: 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).
A: 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?
