How can I show that $\frac{1}{1+|z|^2}$ isn't analytic at $z=1$?

Question

How can I show that $\frac{1}{1+|z|^2}$ isn't analytic at $z=1$? I thought: $\frac{1}{1+|z|^2}=\frac{1}{1+x^2+y^2}$ (substituting $z=x+yi$). We calculate the Cauchy-Riemann equations and we get that $u_x=-2x$, $u_y=-2y$, $v_x=v_y=0$. So this function is only analytic at $z=0$. Is this right?

Answer 1

That fact that it is a real valued function and is not constant tells you that it is not holomorphic.

Consider $\operatorname{f}(x+\operatorname{i}y) = \operatorname{U}(x,y) + \operatorname{i}\operatorname{V}(x,y)$. If $\operatorname{f}$ is real for all $x+\operatorname{i}y$ then $\operatorname{V} \equiv 0$. The Cauchy-Riemann equation $\operatorname{U}_x\equiv\operatorname{V}_y$ and $\operatorname{U}_y \equiv -\operatorname{V}_x$ become $\operatorname{U}_x \equiv 0$ and $\operatorname{U}_y\equiv 0$, i.e. $\operatorname{U}$ is constant.

Answer 2

Analytic (or if you prefer, holomorphic) means it is differentiable on an open set. Your function is only differentiable, i.e. only obeys the Cauchy-Riemann equations, at one point which is not an open set.

Answer 3

Both answers given at this point are great; I'd nevertheless like to suggest another path to victory, based on very little knowledge of the concept of complex-differentiability beyond arithmetic of limits:

If $f(z)=\frac{1}{1+|z|^2}$ were analytic at $z=1$ (that is, there exists some disc $D_r(1)$ about that point in which it's complex-differentiable), then--since it never vanishes--so would $(f(z))^{-1}=1+|z|^2$. Clearly, that implies that $|z|^2=z\bar{z}$ is also analytic in that disc, and since $z\neq 0$ in some (potentially-)smaller disc about $z=1$, so too would $\frac{z\bar{z}}{z}=\bar{z}$ be analytic in that smaller disc. However, one of the most basic facts about complex-differentiability is that $\bar{z}$ is nowhere differentiable.