quotient groups of the complex numbers

Question

I am considering the homomorphism from complex numbers under addition to complex numbers under addition given by

$$f(z)=z+i\overline{z}$$

I need to find the group that $\mathbb{C}/\ker{f}$ is isomorhic to. So I ideally need the image of $f$. I got the following:

$$Im(f) =\{w:w=z+i\overline{z}\}.$$

Letting $w=u+iv$ and $z=x+iy$, this gives

$$u+iv=x+y+i(x+y),$$

i.e. $u=x+y$ and $v=(x+y)$. So $$Im(f) =\{w: u=v\}.$$

But, hoping this is all ok so far, I dont know the name of this group. Can anyone help? In case its needed I got that $$\ker(f)=\{z:z=x-ix\}.$$

Thanks

Answer 1

Note that $Im(f)$ is a line with equation $y = x$. Hence, thanks to a rotation, this is isomorphic to $\mathbb{R}$.

Answer 2

If we write $\;z=x+iy\;$, we have that

$$f(z)=z+i\overline z=x+iy+i(x-iy)=(x+y)+i(x+y)$$

Clearly this is a group homomorphism since

$$f(z+w)=z+w+i(\overline{z+w})=z+i\overline z+w+i\overline w$$

and also

$$\ker f=\left\{\,z=x+iy\in\Bbb C\;:\;x=-y\,\right\}$$

Also, if $\;k\in\Bbb R\;$ , we have

$$f(kz)=kz+i\overline{kz}=kz+ik\overline z=k(z+i\overline z)=kf(z)$$

So in fact $\;f\;$ is a linear transformation $\;f:\Bbb C_{\Bbb R}\to\Bbb C_{\Bbb R}\;$, and thus by the dimension theorem

$$2=\dim_{\Bbb R}\Bbb C=\dim\ker f+\dim\,\text{Im}\,f$$

and since none of the above is trivial, both $\;\ker f,\,\,\text{Im}\,f\;$ are one dimensional real spaces, and thus both isomorphic to $\;\Bbb R\;$