Invariant under transformation $i\mapsto -i$ implies real? When one has an expression in terms of $i$, one can send $i$ to $-i$ and, if the expression remains unchanged, one can conclude that the expression is, in fact, real. Analogous statements hold for expressions involving radicals. Why is this?
One admittedly trivial example is the expression
$$\frac{1}{x-i}+\frac{1}{x+i} .$$
 A: If $x+iy = x-iy$ then $y=0$ (extra characters).
A: This is a basic fact in Galois theory, namely that for Galois extensions $F/K$ the fixed field of the Galois group $G = \text{Gal}(F/K)$ is precisely $K$.  The quadratic extension $\mathbb{Q}(i)/\mathbb{Q}$ is an example of such an extension, as is the quadratic extension $\mathbb{C}/\mathbb{R}$.  (To apply this to your example one can do a few things: either one regards the expression as a power series and applies the above to each term, or one regards the expression as a family of elements of $\mathbb{C}$ and applies the above separately to each of them.)
It is instructive to see how this fails when the extension is not Galois.  A simple example is the extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$.  This extension has no nontrivial automorphisms because the other two roots of $x^3 - 2$ are complex, so the fixed field of the "Galois group" is all of $\mathbb{Q}(\sqrt[3]{2})$.  To get an analogous statement to the one above one has to pass to the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$, which is Galois.
