# prove f(z1)=f(z2) implies z1=z2 and maps to D(0;1)- Complex analysis

• September 19th 2010, 11:36 AM
JJMC89
prove f(z1)=f(z2) implies z1=z2 and maps to D(0;1)- Complex analysis
$f(z)={z\over 1+|z|}$

(i) Prove that $f(z_1)=f(z_2)$ implies $z_1=z_2$.

(iv) Prove that $f$ maps $\mathbb{C}$ onto $D(0;1)$.

[The text suggests using polar coordinates for both of these problems.]
• September 19th 2010, 03:48 PM
tonio
Quote:

Originally Posted by JJMC89
$f(z)={z\over 1+|z|}$

(i) Prove that $f(z_1)=f(z_2)$ implies $z_1=z_2$.

Put $z_1=re^{i\theta}\,,\,\,z_2=se^{i\phi}\,,\,\,r,\,s\ in\mathbb{R}^{+}\,,\,\,0\leq \theta,\,\phi<2\pi$ , so:

$\frac{z_1}{1+|z_1|}=\frac{z_2}{1+|z_2|}\Longrighta rrow \left|\frac{z_1}{1+|z_1|}\right|=\left|\frac{z_2}{ 1+|z_2|}\right|\Longrightarrow \frac{r}{1+r}=\frac{s}{1+s}$ .

From here get that $r=s$ , and then input in the original equality in polar form and get that also $\theta=\phi$ and we're done.

(iv) Prove that $f$ maps $\mathbb{C}$ onto $D(0;1)$.

This should be clear: what's the modulus of.. $\frac{z}{1+|z|}$ ?

Tonio

[The text suggests using polar coordinates for both of these problems.]

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