Show complex function is bijective I'm asked to show that the function $f:\mathbb{C}\setminus\{1\}\rightarrow \mathbb{C}\setminus\{i\}$ given by
$f(z)=i\frac{z+1}{z-1}$ is bijective.
I get stuck very early on - I think I get thrown off by the fact that we're in $\mathbb{C}$ and not good old $\mathbb{R}$.
My attempt:
injective: assuming $f(z_1)=f(z_2)$, I need to show $z_1=z_2$. Usually I would just set it up like this: $$i\frac{z_1+1}{z_1-1}=i\frac{z_2+1}{z_2-1} $$
And then manipulate the expressios to get $z_1=z_2$, but I can't seem to do it here.
Am I choosing a poor strategy here?
In regards to surjectivity, I am just lost.
Any help is much appreciated.
 A: If you solve the equation$$i\frac{z+1}{z-1}=w,$$you will get$$z=\frac{w+i}{w-i}.$$So, define $g\colon\Bbb C\setminus\{i\}\longrightarrow\Bbb C\setminus\{1\}$ by$$g(z)=\frac{z+i}{z-i}$$and check that it is the inverse of $f$. In other words, check that$$(\forall z\in\Bbb C\setminus\{i\}):f\bigl(g(z)\bigr)=z\quad\text{and that}\quad(\forall z\in\Bbb C\setminus\{1\}):g\bigl(f(z)\bigr)=z.$$
A: A broader context: your function is a Möbius transformation $\frac{az+b}{cz+d}$ with$$a=b=c=-d=i\to ad-bc=2\ne0.$$These transformations are famously isomorphic to invertible $\left(\begin{array}{cc}a & b\\c & d\end{array}\right)$.
A: Writing $f(z)$ as
$$
f(z) = i\left (\frac{2}{z-1} + 1\right)
$$
shows that $f$ is the composition of bijective mappings:

*

*$\Bbb{C}\setminus\{1\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto z - 1$,

*$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto \frac{1}{z}$,

*$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{0\} , \, z \mapsto 2 \cdot z $,

*$\Bbb{C}\setminus\{0\} \to \Bbb{C}\setminus\{1\} , \, z \mapsto z + 1$,

*$\Bbb{C}\setminus\{1\} \to \Bbb{C}\setminus\{i\} , \, z \mapsto i \cdot z$
and therefore bijective.
Generally, every Möbius transformation is a composition of simple transformations (translation, inversion/reflection, homothety/rotation), and therefore a bijective mapping from the extended complex plane $\hat{\Bbb C} = \Bbb C \cup \{ \infty \}$ onto itself.
