Is this solution for an homomorphism prove correct? given $f$ : $a+bi$ $; a,b \in  \mathbb{Z}$ ; $ i^2=-1$; is a function over itself defined as $f(a+bi)=a-bi$ ; prove if $f$ is an homomorphism.
in some examples I had seen, it is only to write the functions like this: $(f\cdot g)(x)=f(x) \cdot g(x)$ but I am given a definition of $i^2 = -1$ and I don't know what to do with it
this is my solution, but I am not sure about it:
$f \rightarrow f \rightarrow  \mathbb{Z} $
for  $\forall c,d \in \mathbb{Z}$
a) sum
$(f+f)(a+bi)=(a-bi)+(c-di)$
$(f+f)(a+bi)=f(a-bi)+f(c-di)$
b) product
$(f \cdot f)(a+bi)=(a-bi) \cdot (c-di)$
$(f \cdot f)(a+bi)=f(a-bi) \cdot f(c-di)$
then  $f$ is an homomorphism.
Can anyone with more knowledge in this topic tell me if this is correct or if not, show how can I solve this problem.

Let $X = \{a+bi : a, b \in \mathbb{Z}, i^2 =-1\}$ (although you can just take $\mathbb{C}$ as well with no problem). What you want to show is that $f : X \rightarrow X$ is a homomorphism on the addition and/or multiplication structure. This just means that $f$ preserves the addition and/or multiplication structure (respectively). In other words, given $z, w \in X$ (say $z = a+bi, w = c+di$) you want that
$f(z+w) = f(z) + f(w)$ (preserving the addition structure)
and/or
$f(z\cdot w) = f(z) \cdot f(w)$ (preserving the multiplication structure)
$f(z+w) = f(a+c + (b+d)i) = a+c - (b+d)i = a-bi + c-di = f(z) + f(w)$ so $f$ is an addition preserving homomorphism.
$f(z \cdot w) = f(ac-bd+(ad+bc)i) = ac - bd - (ad + bc)i = (a-bi)(c-di) = f(z) \cdot f(w)$ so $f$ preserves the multiplication structure as well. So $f$ is indeed a homomorphism (in this case a ring homomorphism since it preserves both addition and multiplication).
In general, given a group $(G, +)$, a group homomorphism only needs to preserve the addition structure. Given a ring $(R, + , \cdot)$ a ring homomorphism needs to preserve both the addition and multiplication structures, and so on.
