Why are $i$ and $-i$ "more indistinguishable" than $\sqrt{2}$ and $-\sqrt{2}$? Today I learned that two roots of an irreducible polynomial are "algebraically indistinguishable." 
In $\mathbb{Q}(\sqrt{2})$, define the conjugate of $a+b\sqrt{2}$ as $\overline{a+b\sqrt{2}} = a - b\sqrt{2}$. 
My understanding/intuition on "algebraically indistinguishable" is that if $P$ and $Q$ are algebraic expressions (meaning using the ring operations only) in $\mathbb{Q}(\sqrt{2})$ with $P$ = $Q$, then then we also have $\overline{P} = \overline{Q}$ where $\overline{P}$ and $\overline{Q}$ are the algebraic expressions $P$ and $Q$ except every member of $\mathbb{Q}(\sqrt{2})$ in the expression is replaced with its conjugate.
For example,
$2(3+\sqrt{2})(4+\sqrt{2}) - (6-\sqrt{2}) = 22 + 15\sqrt{2}$
and indeed,
$2(3-\sqrt{2})(4-\sqrt{2}) - (6+\sqrt{2}) = 22 - 15\sqrt{2}$.
And in $\mathbb{R}(i) = \mathbb{C}$, we're guaranteed the same phenomenon:
$2(3+i)(4+i) - (6-i) = 16 + 15i$
$2(3-i)(4-i) - (6+i) = 16 - 15i$
But what I find interesting is that once I involve familiar operations outside of multiplication and addition, this will no longer hold for $\mathbb{Q}(\sqrt{2})$, but it still holds for $\mathbb{R}(i) = \mathbb{C}$.
For example,
$(-\sqrt{2})^{-\sqrt{2}} \neq \overline{\sqrt{2}^{\sqrt{2}}}$
while on the other hand,$(-i)^{-i} = \overline{i^{i}}$.
My initial answer to this is that exponentiation viewed as a binary operation is not closed in $\mathbb{Q}(\sqrt{2})$, so I must view this as an operation on $\mathbb{R}(i) = \mathbb{C}$, but this is a larger field than extending $\mathbb{Q}$ to include the roots of $x^{2}-2$, so in this field with exponentiation, I can now distinguish $\sqrt{2}$ and $-\sqrt{2}$. On the other hand, exponentiation on $\mathbb{R}(i) = \mathbb{C}$ is closed, so even exponentiation does not allow us to algebraically distinguish between $a+bi$ and $a-bi$.
 A: This is such a good question!$\newcommand{\QQ}{\mathbb{Q}}$
My first answer: Let's give a definition of what it means for two elements of a field to be algebraically indistinguishable over a base field. Let $K/L$ be a field extension, and let $\alpha,\beta \in K$. We say
$\alpha$ and $\beta$ are algebraically indistinguishable in $K$ over $L$ if there is an automorphism of $K$ over $L$, $\phi$, such that $\phi(\alpha)=\beta$. (Since any automorphism of a field fixes the characteristic subfield, we will say that $\alpha$ and $\beta$ are algebraically indistinguishable in $K$ if they are algebraically indistinguishable over the characteristic subfield. Since our base field is $\QQ$, which is the characteristic subfield, we'll therefore drop the "over $\QQ$" language).
What does this mean? Well it means that for any polynomial expression in $\alpha$ with coefficients in $L$, say $p(\alpha)$, then $\phi(p(\alpha))=p(\phi(\alpha))$. 
Now because $\QQ(\sqrt{2})$ has an automorphism (conjugation) over $\QQ$ that sends $\sqrt{2}$ to $-\sqrt{2}$, $\sqrt{2}$ and $-\sqrt{2}$ are algebraically indistinguishable in $\QQ(\sqrt{2})$. Similarly $\QQ(i)$ also has an automorphism sending $i$ to $-i$. So $i$ and $-i$ are algebraically indistinguishable in $\QQ(i)$.
Now what if we start taking larger fields? Are $i$ and $-i$ still algebraically indistinguishable as elements of $\newcommand{\CC}{\mathbb{C}}\CC$? Yes! Conjugation is also an automorphism of $\CC$.
How about $\sqrt{2}$ and $-\sqrt{2}$? Are they algebraically indistinguishable in $\newcommand{\RR}{\mathbb{R}}\RR$? in $\CC$? The answer might be a little odd. Note that $\sqrt{2}$ is real, and implicit in our notation is an assumption that $\sqrt{2}$ is positive, and that $-\sqrt{2}$ is negative. Over $\QQ(\sqrt{2})$, we can't tell the difference between the positive and negative square root, but over $\RR$ we can. Every positive element of $\RR$ has a square root in the field, but no negative element does. Thus there cannot be an automorphism $\phi$ of $\RR$ taking $\sqrt{2}$ to $-\sqrt{2}$, since if we had $u^2=\sqrt{2}$, then we would have to have $\phi(u)^2 = -\sqrt{2}$, which is impossible. Thus we can distinguish them over $\RR$. 
However, over $\CC$, we now have all the square roots we could want, and since there is an automorphism of $\QQ(\sqrt{2})$ taking $\sqrt{2}$ to $-\sqrt{2}$, this can be extended to an automorphism of $\CC$ taking $\sqrt{2}$ to $-\sqrt{2}$ via general facts about algebraic closures. So now in $\CC$ we once again cannot distinguish $\QQ(\sqrt{2})$ and $\QQ(-\sqrt{2})$.
My second answer (or rather a continuation): But wait! We're working with $\RR$ and $\CC$! We have a topology. We can talk about other operations than just finite combinations of addition, multiplication and division! So we can define a notion of topologically indistinguishable elements. Let's say that two elements $\alpha,\beta$ of a topological field $K$ of characteristic 0 are topologically indistinguishable if there is a continuous field automorphism $\phi : K\to K$ such that $\phi(\alpha)=\beta$.
From this we get expressions like $\phi(e^\alpha)=e^{\phi(\alpha)} = e^\beta$. More concretely, we know that conjugation from $\CC\to \CC$ is continuous and $\bar{i}=-i$, so by our definition $i$ and $-i$ are topologically indistinguishable. Thus for any expression defined in terms of limits, addition, and multiplication, $F(x_1,x_2,\ldots,x_n)$, 
$$\overline{F(x_1,x_2,\ldots,x_n)} = F(\bar{x_1},\bar{x_2},\ldots,\bar{x_n}).$$
Thus we have $\overline{e^i} = e^{\bar{i}}$.
Now there's a slight complication with the $i^i$ expression, in that $\log i$ is not really well defined. However, for any $w$ such that $e^w = i$, $e^{\bar w} = \bar{i}$, so for any such choice of $w$, $\overline{i^i} =\overline{e^{wi}}= e^{\overline{w}\overline{i}} =\bar{i}^{\bar{i}}$, where the equalities there are really sketchy, since $i^i$ isn't exactly well defined. The central equality is the main point.
So $i$ and $-i$ are topologically equivalent, but what about $\sqrt{2}$ and $-\sqrt{2}$ in $\CC$?
Well, to prove that $\sqrt{2}$ and $-\sqrt{2}$ cannot be topologically equivalent, we just have to find an expression built out of limits, sums, products and divisions on which they differ. One example is
$$\lim_{n\to \infty} (x-1)^n.$$ 
For the positive square root of 2, this expression converges to 0, and for the negative square root of 2 this diverges. Hence there cannot be a continuous field automorphism sending one to the other, since then all such expressions would have to either converge on both $\sqrt{2}$ and $-\sqrt{2}$ or diverge on both.
Thus $\sqrt{2}$ and $-\sqrt{2}$ are not topologically indistinguishable. 
This I believe is the source of what you're observing. Both $\pm i$ and $\pm\sqrt{2}$ are algebraically indistinguishable in $\CC$, but $i$ and $-i$ are also topologically indistinguishable in $\CC$, but $\sqrt{2}$ and $-\sqrt{2}$ are not.
