Originally Posted by

**choovuck** Opalg, I don't like the matrix-valued logarithm, so I slightly modified your idea: instead of taking logarithm and then extending it, we can extend the function f immediately (basically via Schwarz lemma)

for $\displaystyle |z|\ge1$ define $\displaystyle \hat{f}(z)=\left(f(1/\bar{z})^*\right)^{-1}$, which is analytic and well defined since f doesn't vanish. Then on the unit circle $\displaystyle \hat{f}(e^{i\theta})=\left(f(e^{i\theta})^*\right) ^{-1}=f(e^{i\theta})$ by the unitarity condition, so $\displaystyle f=\hat{f}$, i.e. f extends to an entire function. It's bounded, so we're happy.