$f(z)=\frac{z}{1+|z|} $ Is 1-1?

So i have to prove if $f(z_1)=f(z_2)$ then $z_1=z_2$ .

I tried substituting $z=x+yi$ i also tried proving their differences in absolute value are zero.None of this work .Im sure im missing a property or something that will make this very easy. I also tried making the first part of the eq belonging is $R$ so the second part of the eq must have imaginary part zero. I just dont get to the result i want.

Solution 1:

Suppose

$$ \frac{z_{1}}{1+|z_{1}|}=\frac{z_{2}}{1+|z_{2}|}. $$

Then, $$ \frac{|z_{1}|}{1+|z_{1}|}=\frac{|z_{2}|}{1+|z_{2}|} $$ and, hence, $$ |z_{1}|(1+|z_{2}|)=|z_{2}|(1+|z_{1}|)\implies|z_{1}|=|z_{2}|. $$

Thus, from the first equation above, we have $z_{1}=z_{2}$. Thus, $f$ is injective.