Solution 2

Solution 4

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

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

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

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