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?
$|f'(1/2)|≤1/2$
$|f'(1/2)| <4$
$f'(1/2)|≤8$
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
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} $