Is the square root of a complex number over a field always well-defined? A complex number over a field $F$ is defined as $a+b\text{i}$ where $a,b\in F$ and $\text{i}$ is the square root of the inverse of the multiplicative identity of $F$, denoted as $\text{i}^2 = -1$. I have several questions about this definition.


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*Is there any restriction on $F$ in addition to that the square roots of $-1$ must exist? Should $F$ be a quadratically closed field, i.e. every number in $F$ has a square root?


*How square root of a complex number is defined? The square root of a complex number over $\Bbb R$ can be defined by De Moivre's formula, but I don't see how this is extended to an arbitrary field.

For question 2, I can imagine a simple definition. For a complex number $c=\alpha+\beta\text{i}$, we can define the square root of $c$ as a complex number $a+b\text{i}$ st. $a^2-b^2=\alpha,2ab=\beta $. But I don't know if this definition is appropriate.
Any help and reference is appreciated. Thank you!

First, if $F$ already contains something whose square is $-1$ -- for example, if $F=\mathbb C$ or $F=\mathbb F_{37}$. -- then trying to make complex numbers over it goes wrong, and you end up with zero divisors, and not a field.
For example, with complex numbers over $\mathbb F_{37}$ we have $(6+i)(6-i)=0$.
Second, in general there is no canonical way to pick out a single square root of an element of a field. If we have $x$ and some $y$ such that $y^2=x$, then both $y$ and $-y$ will be possible square roots of $x$, and there is no general systematic way to pick exactly one of them to be "the" square root of $x$.
In order to find $y$ given $x$ one can use your computation at the end of the question to find a quadratic equation for $a$ and $b$, and then solve that in $F$ -- which generally requires you to be able to find square roots in $F$ already somehow.
