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Prove that if $f$ is an analytic function in an open set containing the closed unit disk and if $k$ is a positive integer, then there exists a $z_0$ with $|z_0|=1$ and $\left|\frac{1}{z_0^k}-f(z_0)\right|\geq 1$.

Here is what I have: Suppose thios was not true. Then for each $z$ with $|z|=1$, we have $\left|\frac{1}{z^k}-f(z)\right|<1$. From this we get $|1+-z^kf(z)|<|z|^k=1$ on the boundary of the closed unit disk. So by Rouche's theorem, $g(z)=1$ and $-z^kf(z)$ have the same numbers of zeroes in the unit disk. Clearly $-z^kf(z)$ has a zero at 0. So 1 must equal zero somewhere and we have a contradiction.

Is this right? If not, am I on the right track?

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Looks fine to me. –  user31373 Jun 11 '12 at 22:07
    
Looks fine to me. One minor point you haven't stated explicitly is that both functions in Rouche's Theorem must be non-zero on the boundary of the disk. Luckily, the hypothesis precludes this from happening. –  Connor Jun 11 '12 at 22:07
1  
@acoustician Is there ever a need to state that point? Rouche's theorem says that if $|f|>|g|$ on the boundary, then $f$ and $f+g$ have the same number of zeros in the domain. The inequality implies that neither $f$ nor $f+g$ vanish on the boundary. –  user31373 Jun 11 '12 at 23:28
    
I may be missing something, but how does the inequality imply that f+g does not vanish on the boundary? –  Connor Jun 12 '12 at 1:06
    
@acoustician: $|f+g|\geq |f|-|g|$. –  Jonas Meyer Jun 12 '12 at 2:21
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