Complex Arctan function and its power series

I face a sequence of confusing questions:

In complex plane, note that $arctan(z)$ denote the principal branch of inverse complex tanget function ,by requiring $$\frac{-\pi}{2} < \mbox{Re}(\arctan(z))\leq \frac{\pi}{2} .$$

1. Let $g(z) = f(\tan z)$. Show that $g'(z) = 1$ for all $z$ in some domain $D$. Then describe $D$.

I am not sure because we do not know what $g$ actually is. So how can be sure about where $D$ should $g'(z) = 1$. Anyway, I diff it and get $$1= f'(\tan(z))\sec^2 (z)$$ for all $z \in D$. So $f'(\tan(z)) = \cos^2 (z)$ which yileds $f'(z) = \cos^2 (\arctan(z)).$ How to go on form this stage ?

These following 3 questions are connected to this one:

1. Conclude that $f(z) = \arctan (z)$ for $|z| < 1.$
2. Why does the Taylor series for $\arctan$ at the origin not converge in a disc larger than $|z| < 1 ?$
3. Show that $\arctan(1)$ is given by $$\sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$$

Since I stuck at the first one, the following questions do not make much sense for me. Anyway, for 3) I guess that finding radius of convergence might help. Could anyone give suggestions please ?

• @dustin: it's not uncommon to use $\operatorname{Arctan}$ for a particular branch (often the principal branch) of the complex inverse tangent. Same distinction as $\log$ vs. $\operatorname{Log}$. – mrf Feb 17 '15 at 13:48
• @dustin It is as mrf point out. I would like to use the principal branch of inverse function of complex tangent function, $Arctan \ (z).$ – Both Htob Feb 17 '15 at 13:56
• @dustin Alright, I will use $\arctan(z)$ and emphasize that it denotes the principal branch instead. – Both Htob Feb 17 '15 at 14:06
• @dustin it is okay now. I agree that it will be better if I can use valid Latex format than violate it. Anyway, could anyone suggest how to solve the problem ? – Both Htob Feb 17 '15 at 14:11
• You can use \operatorname{Arctan} z to get $\operatorname{Arctan} z$. – mrf Feb 17 '15 at 16:25

From your equation $f'(z) = \cos^2(\arctan(z))$, we have that $$f'(z) = \frac{1}{z^2+1}.$$ If you don't see it, draw a triangle. Integrating with respect to $z$ we get $$f(z) = \arctan(z) + C$$ On the principal domain, $\arctan(\tan(z)) = z$. Therefore, $f(\tan(z)) = z + C$. This should help with $(1)$ and $(2)$. Are you good now?
For $(3)$, consider $$\frac{d}{dz}\arctan(z) = \frac{1}{1 - (-z^2)} = \sum_{n = 0}^{\infty}(-z^2)^n\tag{*}$$ Now, $(*)$ convergences when $$1/R = \limsup_{n\to\infty}\sqrt[n]{\lvert(-z^2)\rvert^n} = \limsup_{n\to\infty} z^2 = \lvert z\rvert^2 < 1$$ That is, when $z$ is in the unit disc. Now you can integrate $(*)$ term by term to get the power series and set $z=0$ to solve for the constant of integration. Now, that you have the power series, plug in $z=1$.
• Okay, I try to make them into $\frac{1}{z^2+1}$. Anyway, I am not sure, I think that using triangle as in real might not a good reason to claim that it is actually $\frac{1}{z^2+1}$ since this is a complex function. – Both Htob Feb 17 '15 at 19:50
• That is real, isn't it ? I do not know I can do that with complex tangent $$\tan(z) = \frac{e^{iz} - e^{-iz}}{i(e^{iz} + e^{-iz})}$$ where $z \in \mathbb{C}$. – Both Htob Feb 17 '15 at 20:04
• Yeah, I know that $\mathbb{R}^2$ is isomorphic to $\mathbb{C}$ as rings. But you sure about that ? That I can do as $\mathbb{R}^2$. If yes, thank you very much for your help. Sorry for asking a lot. – Both Htob Feb 17 '15 at 20:11