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

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

    $\displaystyle f(z)={z\over 1+|z|}$

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

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

    [The text suggests using polar coordinates for both of these problems.]
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  2. #2
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    Quote Originally Posted by JJMC89 View Post
    $\displaystyle f(z)={z\over 1+|z|}$

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


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

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

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


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


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

    Tonio


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