Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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

share|cite|improve this answer

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 $z\in{\mathbb C}\setminus\{\pm i\}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.