Isomorphism in the study of 'local fields' I have a simple question, this is an isomorphism I don't understand
What does the following line mean? This can be interpreted as taking the class of f(z) in the ring $\mathbb{C}[z]/(z-a) \cong \mathbb{C}[z] = \mathbb{C}.$ I don't understand the isomorphism, how and why does this happen?

You have some things wrong. I assume your question is give a $a \in \Bbb{C}$, why is $\Bbb{C}[z]/(z-a) \cong  = \Bbb{C}$?
Well first of all we have the evaluation homomorphism that sends the indeterminate $z$ to $a \in \Bbb{C}$. Since $\Bbb{C}$ is a field, the division algorithm tells us that the kernel of this homomorphism is precisely $(z-a)$. Now the homomorphism is also surjective because for any $w \in \Bbb{C}$, I can write $w = w + a - a$ and so $w$ is the image of the polynomial $z+ (w - a)$.
By the First Isomorphism Theorem it follows that $\Bbb{C}[z]/(z-a) \cong \Bbb{C}$.
