I am trying to find all the $\mathbb{Z}$-submodules of $\mathbb{Z}$ that have a complement
Here with complement I mean:
Def: Let $N \subset M$ be an $R$-Submodule. A submodule $N'$ is called a complement of $N$ if the following two conditions are satisfied \begin{align}N + N' &= M \tag{1} \\ N \cap N' &= \lbrace 0 \rbrace \tag{2} \end{align}
My approach: I know that the $\mathbb{Z}$-submodules of $\mathbb{Z}$ have the form $<m>$ where $m \in \mathbb{Z}$ and $<.>$ denotes the generator. Intuitively I just view $<\lbrace m \rbrace > := <m>$ as 'multiples of $m$' which is indeed a submodule of $\mathbb{Z}$.
I now want to find a complement for such an $<m>$ and I assume that I then have to come up with conditions such that (1) and (2) are satisfied, these should give me the required constrains on $m$, however I do run into trouble.
If $<n>$ for $n \in \mathbb{Z}$ is another $\mathbb{Z}$-submodule of $\mathbb{Z}$ then to satisfy $1$ I must have $m,n \in \mathbb{Z} \setminus \lbrace -1,0,1 \rbrace$ such that $m,n$ are coprime. Then I'd have: $$<m> + <n> = \mathbb{Z} $$ and therefore (1). However if I keep working with said $m,n$ then I only obtain $$<m> \cap <n> = <mn> \tag{*} $$ where I want (*) to be equal to the set only containing zero. If I start with trying to satisfy condition 2 first I'd naturally choose one of the $m,n \in \mathbb{Z}$ to be equal to zero, but then I won't be able to satisfy (1) anymore.
My conclusion is that my entire approach is flawed, since my argumentation runs my into troubles in a cyclic manner.
Sub $\Bbb Z$-modules of $\Bbb Z$ are its deals, however if we have $(n)\cap(m)$ this is always $(\gcd(m,n))\ne \{0\}$ unless $m=0$ or $n=0$.
Note: This is different from what you got, $(2)\cap (4)=(4)\ne (2\cdot 4)=(8)$, for example.
Hence the only complemented sub-modules of $\Bbb Z$ are $\Bbb Z$ and $\{0\}$, and they are complements to each other.