Show $(\mathbb{C},+)/(\mathbb{R},+) \simeq (\mathbb{R},+)$ Show $(\mathbb{C},+)/(\mathbb{R},+) \simeq (\mathbb{R},+)$
Im trying to figure out how the cosets would be. $(a+bi) \in \mathbb{C}, a,b \in \mathbb{R}$ maybe something like $bi+\mathbb{R}$
so maybe $\phi:\mathbb{C}/\mathbb{R} \to \mathbb{R}\\(bi+\mathbb{R}) \mapsto b$ 
1-1:
     $\phi(b_1i+\mathbb{R}) = \phi(b_2i+\mathbb{R}) \implies b_1 = b_2$
or kernel: $\phi(0i + \mathbb{R}) = 0 \implies ker(\phi) = \{0\} \implies \phi: 1-1$
homomorphism: $\phi(bi+\mathbb{R}+ci+\mathbb{R}) = \phi(bi+ci+\mathbb{R}) = b+c = \phi(bi+\mathbb{R}) + \phi(ci+\mathbb{R})  $
onto:
      $\forall a \in R, \exists (ai+\mathbb{R}) $ such that $\phi(ai+\mathbb{R}) = a$, or can say $\phi[ \mathbb{C}/\mathbb{R}] = \{a\, |\, a \in \mathbb{R}\} = \mathbb{R}$ so it is clearly onto.
something like that?
 A: It will probably be easier to define the mapping $\phi:\mathbb{R}\rightarrow \mathbb{C}/\mathbb{R}$ by $\phi(x)= xi+\mathbb{R}$. Then $\ker\phi=\{x\in \mathbb{R}: xi\in \mathbb{R}\}=\{0\}$, so $\phi$ is injective. To show $\phi$ is surjective we need only produce a $y\in \mathbb{R}$, for some $zi +\mathbb{R}\in \mathbb{C}/\Bbb{R}$, such that $\phi(y)= zi+\mathbb{R}$. By our definition we can choose $y=z$. Finally that $\phi$ is a homomorphism follows from $$\phi(x+y)=(x+y)i + \mathbb{R}= xi +yi + \Bbb{R}=xi+\mathbb{R}+yi+\mathbb{R} = \phi(x)+\phi(y)$$
Thus, they are isomorphic.
A: Alternatively, we can display a surjective homomorphism: $\phi: \Bbb C \to \Bbb R$, whose kernel is $\Bbb R$, and apply the Fundamental Isomorphism Theorem.
The homomorphism I have in mind is:
$\phi(x + iy) = y$ (Alternatively, $\phi(z) = \mathfrak{Im}(z)$). It should be clear that $\text{ker }\phi = \Bbb R$, and is surjective. as well ($iy$ is a pre-image of any real $y$).
Geometrically, the cosets are horizontal lines in the complex plane. Since these are totally determined by their imaginary component, this is equivalent to Eoin's answer, without explicit mention of cosets, per se.
In this method, as well, we need not raise the spectre of "well-defined".
