The problem is this: Let $f:\mathbb{R}[x]\rightarrow \mathbb{C}\times \mathbb{C}$ be the homomorphism defined by $f(x)=(1,i)$ and $f(r)=(r,r)$, for $r\in \mathbb{R}.$ Determine the kernel and the image of $f.$
Solution:
Let $K=\ker f=\{p(x)=\sum_{k=0}^na_kx^k\in R[x]: f(p(x))=(0,0)\}$
After a few steps I got: $f(p(x))=\left(\sum_{k=0}^na_k,\sum_{k=0}^n(i)^ka_k\right)=(0,0)$
So then $p(x)$ is in the kernel of $f$ if its coefficients satisfy the following conditions:
$\sum_{k=0}^na_k=0$
$\sum_{k=0}^{\lfloor{n}\rfloor}(-1)^ka_{2k}=0$ $\hspace{1cm}$ and
$\sum_{k=0}^{\lceil{n}\rceil}(-1)^{k-1}a_{2k-1}=0$.
And the image of $f$ would consist of all the pairs (a,b+ic). That is the real part of the first 'coordinate' would be zero.
Am I on the right track, or am I missing something?
Any suggestions are appreciated!