Is $\sqrt{i \cdot \frac{1+z}{1-z}}$ in the Hardy space $H^1(\mathbb{D})$

Question

I am trying to prove that $$\sqrt{i \cdot \frac{1+z}{1-z}}$$ is in the Hardy space $H^1(\mathbb{D})$, where we take the principal branch of the square root. I think that this is true but I am stuck at the moment.

Thanks for the help!

Answer 1

Sure, let $f(z) = \frac{(1+z)^{1/2}}{(1-z)^{1/2}}$ analytic on $|z|<1$ and continuous on $|z|\le 1,z\ne 1$

as $t\to 0$, $f(e^{it}) \sim C t^{-1/2}$

whence $f\in L^1(|z|=1)$.