Prove the Theorem: \[{{i}^{2}}=-1\].

Hint: In order to solve this question, we need to know the meaning of iota and its square, and how it can be simplified to get the required right hand side. Iota is the complex number equivalent to the square root of \[-1\].

Formula used: The formula used here in the question is
\[i=\sqrt{-1}\]
Which we know, that it gives the relation between iota and the real number\[1\], where iota is the complex number as we know.
The other examples of the complex numbers are given as:
\[2i,3i,4i\] and so on are also represented in the form\[\sqrt{-2}\], \[\sqrt{-3}\], \[\sqrt{-4}\]

Complete step-by-step solution:
Proof: Firstly, to solve this question, let us write what the term iota means, iota is a complex number, the value of which is equal to the square root of \[-1\]
i.e.,
\[i=\sqrt{-1}\]
Because the square root of a negative number cannot be evaluated directly because that doesn’t give a real number, so we had to give a notation for this complex number, that notation became iota, \[i\], so that makes iota a complex number
Now if we take the relation given above and square both the sides,
We get, the relation as
\[\begin{align}
  & i=\sqrt{-1} \\
 & \Rightarrow {{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}} \\
 & \Rightarrow {{i}^{2}}=-1 \\
\end{align}\]
So, we get the value of iota square as \[-1\]
Hence, the relation given in the question has been proved.

Note: Alternatively, we can also do this question in the way like
Where the term iota square here is written in the form given and then it is simplified further
\[\begin{align}
  & \Rightarrow {{i}^{2}}=\left( 0,1 \right)\centerdot \left( 0,1 \right) \\
 & \Rightarrow {{i}^{2}}=\left( -1,0 \right) \\
 & \Rightarrow {{i}^{2}}=-1 \\
\end{align}\]
And hence the relation is proved.