Square root of number -1 defined as i, then what is the square root of complex number i?, I would say it should be j as logic suggests but it's not defined in quaternion theory in that way, am I wrong?
EDIT: my question is rather related to nomenclature of definition, while square root of -1 defined as i, why not j defined as square root of i and k square root of j and if those numbers have deeper meanings and usage as in quaternions theory.
Answer
Unfortunately, this cannot be answered definitively. In fact, every non-zero complex number has two distinct square roots, because $-1\ne1,$ but $(-1)^2=1^2.$ When we are discussing real numbers with real square roots, we tend to choose the nonnegative value as "the" default square root, but there is no natural and convenient way to do this when we get outside the real numbers.
In particular, if $j^2=i,$ then putting $j=a+bi$ where $a,b\in\Bbb R,$ we have $$i=j^2=(a+bi)^2=a^2-b^2+2abi,$$ so we need $0=a^2-b^2=(a+b)(a-b)$ and $2ab=1.$ Since $0=(a+b)(a-b),$ then $a=\pm b.$ If we had $a=-b,$ then we would have $1=2ab=-2b^2,$ but this is impossible, since $b$ is real. Hence, we have $a=b,$ so $1=2ab=2b^2,$ whence we have $b=\pm\frac1{\sqrt{2}},$ and so the square roots of $i$ are $\pm\left(\frac1{\sqrt{2}}+\frac1{\sqrt{2}}i\right).$
I discuss in my answer here that $i$ is defined as one of two possible numbers in the complex plane whose square is $-1$ (it doesn't actually matter which, as far as the overall structure of the complex numbers is concerned). Once we've chosen our $i,$ though, we have fixed which "version" of the complex numbers we're talking about. We could then pick a canonical square root of $i$ (and call it $j$), but there's really no point. Once we've picked our number $i,$ we have an algebraically closed field, meaning (incredibly loosely) that we have all the numbers we could want already there, so we can't (or at least don't need to) add more, and there's no particular need to give any others of them special names.
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