# proving the function $\frac{1}{1+x^2}$ is analytic

I have a question, how can we prove that a function, here specificaly the function $\frac{1}{1+x^2}$ is analytic? I know we must show that for any $x_0$ in $\mathbb R$, the series $\sum_{k=0}^{\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ has a convergent radius greater than zero, but how to show that? I appreciate your solutions.

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Showing it directly from the definitions is painful. Why not use some facts about analytic functions? For example, if $f$ is analytic, so is $f^2$, and if $f$ and $g$ are both analytic, then so is $f + g$, etc. –  Zhen Lin Jan 16 '12 at 18:27
As purely a comment, I was recently reading "Visual Complex Analysis," and the Taylor series for this function on the reals came up as a motivation for the natural existence of complex numbers - namely, that the Taylor series for this function around $x_0$ has radius of convergence $\sqrt{x_0^2+1}$. –  Thomas Andrews Jan 16 '12 at 18:40

## 2 Answers

One has $$f(x_0+u)=\frac{\mathrm i}2\left(\frac1{x_0+\mathrm i+u}-\frac1{x_0-\mathrm i+u}\right),$$ hence, for every $|u|\lt|x_0+\mathrm i|=|x_0-\mathrm i|=\sqrt{x_0^2+1}$, $$f(x_0+u)=\frac{\mathrm i}2\sum_{n\geqslant0}(-1)^n\left(\frac1{(x_0+\mathrm i)^{n+1}}-\frac1{(x_0-\mathrm i)^{n+1}}\right)u^n.$$

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It's very simple: The function $f\colon \ x\mapsto{1\over 1+x^2}$ is differentiable with respect to the real variable $x$ on all of ${\mathbb R}$, and the derivative is obtained using the "quotient rule". Now all of this is valid for a complex variable $z$ as well, because for the proof one has used only the laws of algebra and the continuity of the basic field operations, and these are present in ${\mathbb C}$ as well as in ${\mathbb R}$. It follows that $$f'(z)=-{2z\over(1+z^2)^2}$$ for all $z\in{\mathbb C}$ where $f$ is defined, i.e., for all $z\in{\mathbb C}\setminus\{\pm i\}$.

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