I'm facing an exercise that I cannot seem to resolve...
Define the complex function $f(z) = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n+1}}{2n+1}$. Prove that for every $z\in\mathbb{C}$ s.t. $|z|<1$, $tan(f(z))=z$.
The Taylor series of the real $arctan$ function is actually $f$ restricted to the reals, but even if it was a calculus problem I wouldn't know how to prove what is asked. Could someone please point out how to even approach it? I suppose I must differentiate the function, but I don't know how to deal with the power series in complex form.
For $|z|<1$ let $g(z):=\tan(f(z))-z$. For real $x \in (-1,1)$ we know that $g(x)=0$. By the identity theorem we then get $g(z)=0$ for all complex $z$ with $|z|<1$.
Identity theorem: Let $D$ be a region in $ \mathbb C$ and let $f$ and $g$ be holomorphic on $D$. If $(z_n)$ is a convergent sequence in $D$ with $z_n \ne z_m$ for $n \ne m$ , $\lim_{n \to \infty} z_n \in D$ and $f(z_n)=g(z_n)$ for all $n$, then $f=g$ on $D$.