UMD Analysis Qualifying Exam/Aug12 Complex

Solution 4
We know $$|h(z)|=|h(z)-z+z|<R$$ on $$|z|=R$$. Similarly, since $$|h(z)-z|\geq 0$$ then $$|h(z)-z|+|z|\geq R$$ on $$|z|=R$$. This gives $$|h(z)-z+z|<R\leq|h(z)-z|+|z|$$ on $$z=R$$.

So by Rouché's theorem, since both functions are holomorphic (i.e. have no poles), then $$h(z)-z$$ has the same number of zeros as $$z$$ on the domain $$|z|< R$$. Since $$z$$ has only one zero (namely 0), then there is only one solution to $$h(z)=z$$ inside the open disc $$|z|< R$$.

Observe that $$ h(z) \neq z $$ for any $$ |z| = R $$, since that would imply $$ |h(z)| = R $$ for some $$ z $$ on the boundary, contradicting the hypothesis.

Thus, there is only one solution to $$h(z)=z$$ inside the open disc $$|z| \leq R$$.