Identity theorem. Calculate $f$ using $f'$ Supose $f\in\mathcal{H}(D(0,2))$. If $f'(1/n) = \frac{1}{n} e^{1/n^2}$, calculate $f(i) - f(-i)$. 
Anyone can help me? Thank you.
My first attemp was prove that $f'(z)=ze^{z^2}$ in $D(0,2)$ using the identity theorem. 
If $w_n = \frac{1}{n} \in D(0,2)$, it verifies: 
$$\left\lbrace \begin{array}{l}  \lim_{n \rightarrow \infty} w_n = 0 \\ 0\in D(0,2) \\ w_n \not= 0 \ \forall n \end{array} \right.$$
If $g(z) = ze^{z^2}$, it verifies that $f'(1/n) = g(1/n)$. 
Using the identity theorem, follows $$f'(z) = g(z) \  \forall z \in D(0,2)$$
But I don't know how to calculate $f(i)-f(-i)$ using $f'(z)$
 A: You have proved that $f'(z)=ze^{z^2}$.
This means that $f(z)=\frac12e^{z^2}+C$ for some complex constant $C$.
(Just like in analysis over the reals, it's enough to guess one integral function, as integral functions are unique up to additive constants.)
You don't know what $C$ is, but you don't need to, since it cancels out:
$$
f(i)-f(-i)
=
(\frac12e^{i^2}+C)
-
(\frac12e^{(-i)^2}+C)
=
\frac12e^{-1}
-
\frac12e^{-1}
=
0.
$$
Alternatively, since your function is analytic, you could calculate the path integral of the derivative from $-i$ to $i$ to get the difference of the values.
In that approach the constant never even appears.
More details on this alternative approach:
You can define a function $g:\mathbb R\to\mathbb C$ by $g(t)=f(it)$.
You can find $g'(t)$ in terms of $f'$ and the use the fundamental theorem of calculus to find $g(1)-g(-1)=\int_{-1}^1g'(t)dt$.
In fact, any path between the two points will do, but integrating along the imaginary axis will be easiest.
For more details, please ask a new question or see this question on the fundamental theorem of calculus in the complex plane.
