I have this function $q:\mathbb{H}\rightarrow\mathbb{C}^-, z\mapsto -z^2$, with $\mathbb{H}$ the upper half plane, i.e. $\mathbb{H}=\{z=x+iy|y>0\}$ and $\mathbb{C}^-=\mathbb{C}-\mathbb{R}_{\leq 0}$, the open set obtained by deleting the negative real part.
I already proved that $q$ is holomorphic and bijective (okay if I'm honest, I only proved the injective part, I still have to find out how to prove the surjective part) . Now I want to use the fact that holomorphic functions are analytic, and if $q$ is analytic and injective and $q(\mathbb{H})=\mathbb{C}^-$, then $q:\mathbb{H}\rightarrow \mathbb{C}^-$ is an analytic isomorphism. However, I'm not sure how to prove that $q(\mathbb{H})=\mathbb{C}^-$. is this true because it is surjective? But how do I prove it is surjective?
Hope someone can help me :)
Note that $$ \mathbb H=\{r\mathrm{e}^{i\vartheta}: r>0\,\,\&\,\,0<\vartheta<\pi\}, \quad\mathbb C^-=\{r\mathrm{e}^{i\vartheta}: r>0\,\,\&\,\,-\pi<\vartheta<\pi\}. $$ So, if $f(z)=-z^2$, then $f(r,\vartheta)=(r^2,2\vartheta-\pi).$
It becomes thus much easier to check that $f$ is bijective.