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 $$\frac{-\pi}{2} < \mbox{Re}(\arctan(z))\leq \frac{\pi}{2} .$$
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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:
- Conclude that $f(z) = \arctan (z)$ for $|z| < 1.$
- Why does the Taylor series for $\arctan$ at the origin not converge in a disc larger than $|z| < 1 ?$
- 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 ?
Best Answer
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 $z=1$.