Find **Pontrjagin dual.** of $\Bbb Z_n$ & $\Bbb Z$ 
Let $G$ be a group.  Consider the set $Hom(G,\Bbb C^*)$ of homomorphisms from $G$ to $\Bbb C^*$.
Define a binary operation $+:Hom(G,\Bbb C^*)\times Hom(G,\Bbb C^*)\to Hom(G,\Bbb C^*)$ s.t 
  $$(f_1+f_2)(g)=(f_1 (g))(f_2 (g)).$$
This group $(Hom(G,\Bbb C^*),+)$ is called the Pontrjagin dual.

Now my questions are:
1) What is $ Hom(\Bbb Z_n,\Bbb C^*)$?? 
2) What is $ Hom(\Bbb Z,\Bbb C^*)$??

My intuition is:

1) we can construct $\phi:\Bbb Z_n \to \Bbb C^*$ then $Im(\phi)\le n $ 
Again $\Bbb Z_n/{\ker \phi} \cong Im(\phi)$.
Some finite subgroups of $\Bbb C^*$ are in $S^1$, i.e., 
$\phi:\Bbb Z_n \to S^1$ s.t $x \mapsto e^{2x\frac{\pi}{n}} $, if $n$ is not prime then we can construct more like this also in each case we have the trivial homomorphism.
So are these the only possibilities or there are more?? Again prove whatever your conclusion is.
2) Same like 1) here we can construct $\phi:\Bbb Z \to S^1$ s.t $x \mapsto e^{2x\frac{\pi}{r}} $ , where $r \in \Bbb R \setminus \Bbb Q$. What next now??
 A: If $\phi: \Bbb{Z}_n \to \Bbb{C}^*$, put $z = \phi(1) \in \Bbb{C}^*$.  Then,
\begin{align}
z^n &= \underbrace{\phi(1) \cdot \cdots \cdot \phi(1)}_{n} \\
&= \underbrace{(\phi + \cdots + \phi)}_{n}(1) \\
&= (n\phi)(1) \\
&= \operatorname{id}_{\Bbb{Z}_n}(1) \\
&= 1,
\end{align}
so $z$ is an $n$-th root of unity.  By the fundamental theorem of algebra, there are exactly $n$ of these.  If $\zeta$ is a primitive $n$-th root of unity, then $z = \zeta^k$ for some $k \in \{0, 1, \ldots, n-1\}$.  This parametrizes possible values of $z$, so the dual is a group of size $n$.
It turns out that the dual group is cyclic, so
$$
\operatorname{Hom}(\Bbb{Z}_n, \Bbb{C}^*) \cong \Bbb{Z}_n.
$$
Why?  Say that $\psi: \Bbb{Z}_n \to \Bbb{C}^*$ with $\psi(1) = \zeta$, a primitive $n$-th root of unity.  Then, the following calculation shows that any other $\phi: \Bbb{Z}_n \to \Bbb{C}^*$ is a multiple of $\psi$:
$$
\phi(1) = \zeta^k = \bigl( \psi(1) \bigr)^k = (k\psi)(1),
$$
and any homomorphism from $\Bbb{Z}_n$ is uniquely determined by where it sends $1$.

The case of $\operatorname{Hom}(\Bbb{Z}, \Bbb{C}^*)$ is easier.  Any homomorphism $\phi: \Bbb{Z} \to \Bbb{C}^*$ is uniquely determined by $z = \phi(1)$.  Unlike the case of the finite cyclic group, there are no restrictions on $z \in \Bbb{C}^*$.  Thus,
$$
\operatorname{Hom}(\Bbb{Z}, \Bbb{C}^*) \cong \bigl( \Bbb{C}^*, \times \bigr).
$$
