If $f$ is defined on the Upper-Half Plane and $f(in)=e^{-n},$ find $f(1+i)$ 
Let $f$ be a bounded analytic function in the upper half plane. Suppose that $f(in)=e^{-n}$ for all $n\in \mathbb{N}$. Find $f(1+i)$ and explain why the value you found is the only one possible.

I don't know how to proceed. I have two ideas, but I don't think they have any hope.
Let $g(z)=f(z)-e^{-z}$. Then $g(in)=0$ for $n\in \mathbb{N}$. I cannot apply the identity theorem to $g$ since the limit point of its zero set is $\infty$ which doesn't belong to the upper half-plane. 
The other idea was to apply Cayley's transformation $\frac{z-i}{z+i}$ to $g$, but in this case the limit point becomes $1$, which doesn't belong to the boundary of the unit disc. In either case, the identity theorem cannot be applied. Any input would be greatly appreciated.
 A: First of all, you have a small typo in your definition of $g$. Put $g(z)=f(z)-e^{iz}$. Then $g(in)=0$ for all $n\in\mathbb{N}$. Furthermore $g$ is bounded on the upper half plane, since
$$|e^{iz}| = |e^{i(x+iy)}| = e^{-y} \le 1. $$
What remains is to exploit the boundedness of $g$. The zeros of a bounded non-zero holomorphic function satisfies Blaschke's condition. If you transfer that condition to the upper half plane (see for example this) you get
$$ \sum_{j=1}^{\infty} \operatorname{Im}\bigg( \frac1{\alpha_j} \bigg) < \infty $$
where $\alpha_j$ are the zeros. In your example, the series diverges, which forces $g$ to vanish identically. Hence $f(1+i)=e^{i(1+i)}$.
As an exercise, show that there are other unbounded functions satisifying the conditions.

More details: The above isn't quite the full story. For the disc, Blaschke's condition is $\sum_j (1-|\alpha_j|) < \infty$. Using the Cayley transform $z \mapsto (z-i)/(z+i)$ indeed transforms the condition on the upper half-plane to
$$
\sum_j 1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big| < \infty.
$$
For the specific question, we can directly check that
$$
\sum_j 1 - \Big| \frac{ij-i}{ij+i} \Big| =
\sum_j 1 - \Big| \frac{j-1}{j+1} \Big| =
\sum_j \frac{2}{j+1} = \infty.
$$
For the general case, the condition I posted originally is nicer than the one above, but we can check that they are almost equivalent:
First of all,
$$
1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big|^2 = 
\frac{|\alpha_j+i|^2 - |\alpha_j-i|^2}{|\alpha_j+i|}^2 =
\frac{4\operatorname{Im} \alpha_j}{|\alpha_j+i|^2}
$$
and since
$$
1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big|^2 =
\bigg( 1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big| \bigg)
\bigg( 1 + \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big| \bigg)
$$
we see that
$$
1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big| \le
1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big|^2 \le
4\bigg( 1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big| \bigg)
$$
so our series converges if and only if
$$
\sum_j 1 - \Big| \frac{\alpha_j-i}{\alpha_j+i} \Big|^2 = 
\sum_j \frac{4\operatorname{Im} \alpha_j}{|\alpha_j+i|^2} 
$$
does.
Now, this last series converges if and only if:
$$
\sum_{|\alpha_j| < 1} \operatorname{Im}(\alpha_j) < \infty
\quad\text{and}\quad
\sum_{|\alpha_j|\ge 1} \operatorname{Im}\bigg( \frac1{\alpha_j} \bigg) < \infty, 
$$
so for example, if there are only finitely many zeros in the unit disc, the condition I posted originally is enough.
