Can an Entire Function with Given Conditions Have a Zero in a Specific Region?

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If f is entire, and 1<=\left|f\right| <=2 for all \left|z\right| =1, and there is a z0 with \left|z0\right| <1 and f(z0)=z0, then prove or disprove that there exist a z1 with \left|z1\right|<1 such that f(z1)=0.
 
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I am thinking of using Rouche's Theorem, but find it not that useful.
 
a bounded entire function is a constant
 
wofsy said:
a bounded entire function is a constant

This is only true if it was bounded on the entire complex-plane, here f is not even bounded in a domain, but on a line.
 
elibj123 said:
This is only true if it was bounded on the entire complex-plane, here f is not even bounded in a domain, but on a line.

right I misread your problem
 
huyichen said:
If f is entire, and 1<=\left|f\right| <=2 for all \left|z\right| =1, and there is a z0 with \left|z0\right| <1 and f(z0)=z0, then prove or disprove that there exist a z1 with \left|z1\right|<1 such that f(z1)=0.

I think your idea of using Rouche' Theorem is right.

If z0 = 0 there is nothing to prove.
If z0 is not zero then

let g(z) = f(z) - z0. |f - g| = |z0| < 1 <= min(f(z) on the unit circle.

So f and g have the same number of roots inside the unit circle ( f has no roots on the unit circle.)

I am not sure why you need f to be entire.
 
That's right, thanks!
 
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