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let f be an analytic function and inside positivity oriented unit circle -y such that $|f(z)-z|<1$ on y then.

which of the following is correct?

  1. $|f'(1/2)|≤1/2$

  2. $|f'(1/2)| <4$

  3. $f'(1/2)|≤8$

4) f has at least one zero in $C$

By Schwarz lemma I can say that $f(z) = az$ where $a<1$ and $f'(z) < 1$ so my answer is option 1 and 4

Is it correct pliz tell me....

Any hints or solution can be appreciated

Thanks u

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    $\begingroup$ It is true that $f$ has at least one zero inside the disc, by the argument principle. On the other hand, the function $f(z)=(2-\epsilon)z$, for small $\epsilon>0$ satisfies $|f(z)-z|=(1-\epsilon)|z|<1$ on the unit circle, and $|f'(1/2)|=2-\epsilon>1/2$. The inequality tells you that $|f(z)|<2$ on the unit circle. You can then try to apply Schwarz's lemma to $f/2$. $\endgroup$ – YAlexandrov Mar 11 '18 at 10:20
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    $\begingroup$ Applying Schwarz's, the more flexible version I am getting the bound $2/3$, which tells that 2 and 3 hold, but I just came back from flying around earth/2 so you better check it. $\endgroup$ – YAlexandrov Mar 11 '18 at 10:32
  • $\begingroup$ i m getting only option 4 correct by f/2 @YAlexandrov $\endgroup$ – user525416 Mar 11 '18 at 10:37
  • $\begingroup$ is its correct@YAlexandrov $\endgroup$ – user525416 Mar 11 '18 at 10:42
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I am assuming you mean that $|f(z) -z | < 1$ for all z such that $ |z| =1 $

This is equivalent to $ |f(z) -z | < |z| $ for all z such that $ |z| =1 $

You can then apply Rouche's theorem with $g(z) = z$ and $h(z) = f(z)-z$, together with the above property you get that $h+g$ and $ g$ have the same number of zeros in the unit disc. Namely, -$h+g(z) = f(z)-z +z = f(z)$ has a zero in the disc since clearly g has a zero at 0.

1 is not correct since you can use $f(z) = \frac{3}{2}z $ . f has the property $\forall z $ such that |z| =1 $ |f(z) -z | = |\frac{1}{2}z| < 1 $ , but
$f'(\frac{1}{2}) = \frac{3}{2} > \frac{1}{2} $

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  • $\begingroup$ thanks u so much @Daphna keidar $\endgroup$ – user525416 Mar 15 '18 at 0:09

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