Let $\Omega=\mathbb{C}\setminus K$ where $K$ is a compact connected set containing $1+i$, $1-i$, $−1+i$, and $−1−i$. I want to prove that there exists a holomorphic function $f:\Omega\to\mathbb{C}$ such that $(f(z))^4=z^4+4$. In other words, I am trying to show $z^4+4$ has a holomorphic $4$th root on $\Omega$.
What I was able to confirm immediately is that the image of $\Omega$ under $z^4+4$ cannot contain $0$ due to the values we excluded from $\Omega$. However, it is not necessarily the case that $\Omega$ is simply connected, or even connected, so I can't invoke the Monodromy theorem. Any ideas on what to do?
You are right that $\Omega$ may not be connected, but it suffices to show that such a function $f$ exists in every connected component $U$ of $\Omega$.
Let $\{\omega_1,\omega_2,\omega_3,\omega_4 \} = \{ \pm 1 \pm i \} $ be the roots of $$ p(z) = z^4+4 = (z-\omega_1)(z-\omega_2)(z-\omega_3)(z-\omega_4) \, . $$ Now let $\gamma$ be an arbitrary closed path in $U$. The winding number $$ I(\gamma, a) = \frac{1}{2\pi i}\int_{\gamma} \frac{dz}{z-a} $$ is integer-valued and (as a function of $a$) constant in each connected component of $\Bbb C \setminus U$. Since $K$ is connected and contains all $\omega_j$ it follows that $$ I(\gamma, \omega_1) = I(\gamma, \omega_2) = I(\gamma, \omega_3) = I(\gamma, \omega_4)\, , $$ so that $$ \tag{*} \frac 14 \int_\gamma \frac{p'(z)}{p(z)} \, dz = 2 \pi i \frac 14 \sum_{j=1}^4 I(\gamma, \omega_j) = 2 \pi i k $$ for some integer $k$. Therefore we can fix a point $z_0 \in U$ and define $f : U \to \Bbb C$ as $$ f(z) = \exp \left( \frac 14 \int_{z_0}^z \frac{p'(z)}{p(z)} \, dz\right) $$ where the integral is taken along any path from $z_0$ to $z$ in $U$. $(*)$ shows that this definition is independent of the choice of the integration path. Then $$ 4 \frac{f'(z)}{f(z)} = \frac{p'(z)}{p(z)} $$ in $U$, so that $f(z)^4 = Cp(z)$ with some constant $C \in \Bbb C$. Finally multiply $f$ with a suitable constant so that $f(z)^4 = p(z)$ in $U$.