# Complex Variable Problem

1. Jan 11, 2010

### huyichen

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.

2. Jan 11, 2010

### huyichen

I am thinking of using Rouche's Theorem, but find it not that useful.

3. Jan 12, 2010

### wofsy

a bounded entire function is a constant

4. Jan 12, 2010

### elibj123

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.

5. Jan 12, 2010

### wofsy

6. Jan 12, 2010

### wofsy

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.

7. Jan 13, 2010

### huyichen

That's right, thanks!