Ramification in $\mathbb Q(i,\sqrt[4]\pi)/\mathbb Q(i)$

I find this question somewhat frustrating, ’cause I haven’t been able to answer it to a degree of completeness that satisfies me, but at least I can help. I don’t see any way of doing this in a few lines.

First, since you’re interested only in the ramification and splitting of the prime $\mathfrak m=(1+i)$ in your extension $K=k(\pi^{1/4})$, where $k=\Bbb Q(i)$, the question is purely local, and we can feel free to localize and complete and thus work over $k'=\Bbb Q_2(i)$.

Second, I’m sure you know that since the extension $K\supset k$ is Galois, all the primes above $\mathfrak m$ have the same ramification index and residue field degree. So we have $efg=4$, where as usual, $e$ is the ramification index, $f$ is the residue field degree, and $g$ is the number of primes of $K$ above $\mathfrak m$. So there are apparently six cases: $e=4$ (totally ramified), $f=4$ (inert), $g=4$ (totally split), and the three mixed cases $e=f=2$, $e=g=2$, and $f=g=2$. Since the extension is cyclic, all six occur.

Third, we’re working with a Kummer extension, and we can use the facts and techniques of Kummer theory.

We could get to work right now on your problem, but I think it’s useful to look at two simpler analogous situations. If your base is $\Bbb Q$ instead, and ask about the splitting and ramification above $2$ of the extension $\Bbb Q(\sqrt n)$, where $n$ is odd (in all this, primality of your $\pi$ is not significant), then the answer depends entirely on the congruence of $n$ modulo $8$: if $n\equiv1$, then $2$ splits; if $n\equiv3$ or $7$, then $2$ ramifies; and if $n\equiv5$, the prime $2$ remains prime, i.e. is inert. You can see this very easily by passing to the complete situation, and ask about what happens when you adjoin $\sqrt m$ to $\Bbb Z_2$. Kummer extension, the quadratic extensions are told to you by $\Bbb Q_2^*/{\Bbb Q_2^*}^2$. Since you’re interested only in the square roots of odd numbers, you're reduced to looking at $\Bbb Z_2^*/{\Bbb Z_2^*}^2$. As you probably know, the multiplicative structure of $\Bbb Z_2^*$ is isomorphic to the additive group $C_2\oplus\Bbb Z_2$, where $C_m$ is the cyclic group of order $m$. The two generators are $-1$ and $5$. The squares here are therefore of index $4$, and you easily see that the subgroup $1+8\Bbb Z_2$ is contained in the squares but also has index $4$, so that ${\Bbb Z_2^*}^2=1+8\Bbb Z_2$ The four representatives of the quotient are $\{\pm1,\pm3\}$, corresponding to the four cases I mentioned above.

Let’s do the same for quadratic extensions of $k'=\Bbb Q_2(i)$ generated by the square roots of “odd” elements, that is numbers $\pi$ not divisible by $1+i$. I’ll call the ring of integers here $\mathfrak o$, equal to $\Bbb Z_2[\varpi]$, where $\varpi=1+i$ is a prime element in the local ring $\mathfrak o$; I’ll call the ideal that $\varpi$ generates $\mathfrak m$ again. Now, since the residue field is just $\Bbb F_2$, the units of $\mathfrak o$ are again $1+\mathfrak m$, which now has a multiplicative structure isomorphic to the additive group $C_4\oplus\Bbb Z_2\oplus\Bbb Z_2$. The squares therefore are of index $8$, and again $1+4\mathfrak m$ is contained in the squares, but has index $16$; the nonsquare not in $1+4\mathfrak m$ is $-1$, so that the squares are $\langle-1,1+4\mathfrak m\rangle$. So the splitting and ramification of $\mathfrak m$ in $K=k(\sqrt\pi)$ depends entirely on the congruence of $\pi$ modulo the multiplicative subgroup $\langle-1,1+4\mathfrak m\rangle$ of $1+\mathfrak m=\mathfrak o^*$. For instance, if $\pi\equiv1$, i.e. if $\pi\in{\mathfrak o^*}^2$, then $(1+i)$ splits; if $\pi\equiv5$, $(1+i)$ remains inert in the extension; and in all other cases, $(1+i)$ ramifies.

Finally, the question that you did ask. Now, because we’re working with a Kummer extension of degree $4$, you need to worry about ${k^*}^4$ as a subgroup of $k^*$, but because of your restriction to the fourth roots of odd Gaussian numbers, you need to examine $\mathfrak o^*=1+\mathfrak m$ and the subroup of its fourth powers. Now the subgroup $1+8\mathfrak m\subset1+\mathfrak m$ happens to be equal to ${\mathfrak o^*}^4$, the index being $64$. Again, if $\pi$ lands on the identity element of the quotient group, i.e. if $\pi\equiv1\pmod{1+8\mathfrak m}$, then $(1+i)$ splits completely in the extension. You can try any particular one of the other $63$ classes to see what the behavior is, depending on how the polynomial $X^4-\pi$ splits over $k'$. Because there are so many mixed cases, though, I don’t know what sensible general statements one may make. And I feel bad that I haven’t identified the (unique) class in the quotient group that gives you the inert case, except to check that it’s not $\sqrt[4]5\,$: that one gives a mixed case, $e=1$, $f=g=2$, as I recall.