This question is related to the computation suggested here: Finding the subfields of the cyclotomic field of order $5$
Let $K = \Bbb{Q(\zeta)}/\Bbb{Q}$ where $\zeta$ is a primitive 5th root of unity. We can identify the Galois group with the following 4 automorphisms:
$$ \mathrm{id}: \zeta \mapsto \zeta \\ \tau_2: \zeta \mapsto \zeta^2 \\ \tau_3: \zeta \mapsto \zeta^3 \\ \tau_4: \zeta \mapsto \zeta^4 \\ $$
A quick check shows that the order of $\tau_4$ is 2, so I'd like to compute its fixed field $H$.
Letting $x$ be an arbitrary element of $K$, I've computed its image under $\tau_4$:
$$ x = a + b\zeta + c \zeta^2 + d\zeta^3 + e\zeta^4 \quad \mapsto \quad \tau_4(x) = a + e\zeta + d\zeta^2 + c\zeta^3 + b\zeta^4. $$
And so if $x=\tau_4(x)$, then $b=e$ and $d=c$, so $$ x = a + t_1(\zeta + \zeta^4) + t_2(\zeta^2 + \zeta^3) $$
for some $t_1, t_2 \in \Bbb{Q}$.
It seems like the actual fixed field is $\Bbb{Q}(\zeta + \zeta^4)$, but I'm not sure if I can conclude that from what I've written.
Have I messed up a calculation somewhere? Or is it perhaps the case that $\zeta^2 + \zeta^3 \in \Bbb{Q}(\zeta + \zeta^4)$?