How to prove that f(z) is univalent?

Question

$f(z) =z/1-z^2$

I know, i'll have to show that $f(z_1) = f(z_2) $ For the case of $f(z) =z^2$, it is quite straight forward: $z_1^2 - z_2^2 = 0$

$(z_1-z_2)(z_1+z_2) = 0$

This gives

$z_1-z_2 = 0$

$z_1 = z_2$ QED

How do i manipulate $f(z) =z/1-z^2$ in such a way,to show that is univalent?

Answer 1

$f$ is univalent in region $D$ (not containg $\pm 1$) of $\mathbb C$ iff the following conclusion holds:

$z \in D$ implies ($z=0$ or $z \neq 0$ and) $-\frac 1 z \notin D$ except when $z=i$.

Examples of regions $D$ satisfying this condition are $\{z:|z|<1\}$ and $\{z: |z| >1\}$.

Proof of the equivalence: if $D$ satisfies this condition then $\frac a {1-a^{2}}=\frac b {1-b^{2}}$ implies $a-ab^{2}=b-a^{2}b$ which implies $(a-b)=-ab(a-b)$ Since $ab \neq -1$ it follows that $a=b$. Conversely. suppose $D$ does not satisfy the stated condition. Then there exist $a,b \in D$ such that $ab=-1$ and $a \neq i$. You can now check that $f(a)=f(b)$ and $a\neq b$.

Answer 2

I assume that $f(z)=\frac{z}{1-z^2}$ for $z \ne \pm 1$.

If $z_1 \ne \pm1$ and $z_2= - \frac{1}{z_1}$, then $f(z_1)=f(z_2).$

Conclusion: $f$ is not univalent.