Let $\bar{\mathbb Q}$ be an algebraic closure of $\mathbb Q$. Determine all homomorphisms from $\mathbb Q(\sqrt[4]{2},i)\rightarrow\bar{\mathbb Q}$ and their images!
Now the minimal polynomials for $\sqrt[4]{2}$ is $x^4-2$ and for $i$ it's $x^2+1$. So if $\sigma$ is a homomorphism from the rationals to a closure of them, its the identity as $\sigma(1)=1$. So now I add $i$ so I get $\sigma':\mathbb Q(i)\rightarrow\bar{\mathbb Q}$, extending $\sigma$, so for $\sigma'(i)$ I got 2 possibilitys, namely the two roots of $i$'s minimal polynommial, $i,-i$. Similary if I extend further I get $4$ possibilitys for $\sqrt[4]{2}$, namely $\sqrt[4]{2},-\sqrt[4]{2},i\sqrt[4]{2},-i\sqrt[4]{2}$. So is it correct to conclude that there are $8$ different homomorphisms possible? Now for their images I've got 2 different images, depending on if you chose $\pm \sqrt[4]{2}$ or $\pm i\sqrt[4]{2}$ as image of $\sqrt[4]{2}$, namely $a+b\sqrt[4]{2}+c\sqrt{2}+d2^{3/4}+ei$ for first and for the later $a+bi\sqrt[4]{2}+c\sqrt{2}+di2^{3/4}+ei$, $a,b,c,d,e\in\mathbb Q$. Is this correct?
Your reasoning and counting of possible images of $i$ and $\sqrt[4]2$ (and thus of homomorphisms in total) is correct. However, your description of the images is not. For example, the image will always contain $i\sqrt[4]2$, if not as $\sigma(\sqrt[4]2)$ directly then as product (or negative product) of $\sigma(\sqrt[4]2)$ and $\sigma(i)$. You should notice that theimage is always the same - and is eight-dimensional.