We have to show $\mathbb{C}[x]/(x-i) \cong \mathbb{C}$
Lets consider $\psi: \mathbb{C}[x] \to \mathbb{C}$ define $f(x) \mapsto f(i)$. Then $\psi$ is ring homomorphism as $\psi(f(x)+g(x))=f(i)+g(i)=\psi(f(x)+\psi(g(x)) \ \text{and} \ \psi(f(x)(g(x))=f(i)g(i)=\psi(f(x))\psi(g(x)) \ \text{and} \ \psi(1)=1$
$\psi$ is onto as if $a+ib \in \mathbb{C}$ then there exists $f(x)=a+xb \in \mathbb{C}[x]$ suchthat $\psi(f(x))=f(i)$
Now $ker \psi=\{f(x) \in \mathbb{C}[x]| f(i)=0\}=\{f(x) \in \mathbb{C}[x]| \text{now} \ f(x)=(x-i)q(x)+f(i) \ \text{but} \ f(i)=0 \ \text{so} \ f(x)=(x-i)q(x)\}=\{f(x) \in \mathbb{C}[x]| f(x)=(x-i)q(x)\}=(x-i)$
So from first isomorphism theorem $$\mathbb{C}[x]/(x-i) \cong \mathbb{C}$$
Now to show $\mathbb{C}[x]/(x+i) \cong \mathbb{C}$ we just take the map $f(x) \mapsto f(\bar i)$ where $\bar i=-i$and all the process is same as just above. Is it okay?
Looks ok, except for you did not define $a,b$ in the second paragraph. I would improve it like following:
$\psi$ is onto $\mathbb{C}$ because for any $z \in \mathbb{C}$ there exists (a constant function) $f(x)=z \in \mathbb{C}[x]$ satisfying $\psi(f(x))=z$