The function $\exp(\frac{z-1}{z+1})$

Question

Let $D=\{z\in\mathbb{C}:|z|<1\}$. Consider the function $f:D\longrightarrow \mathbb{C}$ defined as $f(z)=\exp(\frac{z-1}{z+1})$. Example 1.3.7 in page 21 of the book Interpolation, Identification, and Sampling By Jonathan Richard Partington says the function $f$ belongs to the hardy space $H^{\infty}$ but not in the disc algebra $A(\bar D)$. Can any one tell why? So basically

1. why is the function holomorphic on $D$?
2. why is it bounded on $D$?
3. Why is it not continuous on the boundary $\{z\in \mathbb{C}: |z|=1\}$.?

Answer 1

1. The function is holomorphic because it is the composition of holomorphic functions. Observe that the only singular point is $z=-1\notin D$.
2. The function $(z-1)/(1+z)$ is a Linear Fractional Transformation, also known as a Möbius function. It takes $D$ onto the left hand half-plane $H=\{z:\Re(z)<0\}$. The exponential function is bounded on $H$, because $$ \bigl|e^{x+iy}\bigr|=e^x\le1\text{ if }x\le0. $$
3. If $\theta\in(-\pi,\pi]$, then $$ \exp\Bigl(\frac{1-e^{i\theta}}{1+e^{i\theta}}\Bigr)=\cos\Bigl(\tan\frac{\theta}{2}\Bigr)-i\sin\Bigl(\tan\frac{\theta}{2}\Bigr). $$ This function is not continuous at $\theta=\pi$.