# $f(z)=\frac 1 {x^2+y^2}+i \frac 1 {x^2+y^2}$ is differentiable and holomorphic

Is the function $f(z)=\dfrac 1 {x^2+y^2} + i \dfrac 1 {x^2+y^2}$ is differentiable and holomorphic somewhere?

We have $z=x+iy$ and $f(z)= \dfrac{1+i}{|z|^2}$ . Now, $f(0)= \lim_{\delta z \rightarrow 0} \frac{f(0+\delta z) - f(0)}{\delta z}=$ undefined. So $f(z)$ is not differentiable at origin. Also since since $f(z)= |z|^2$ is not differentiable anywhere except origin , so $f(z)$ is not not differentiable and hence it is not holomorphic. But I am not sure, please help me

• One should not write "$=\text{undefined}$". Rather, one should say that something is undefined. $\qquad$ – Michael Hardy Apr 10 '17 at 21:37

## 2 Answers

You can check for sure using the Cauchy-Riemann equations. If we have a complex function $f(x+yi) = u(x,y) + iv(x,y)$, then for $f$ to be differentiable at a point $z_0 = x_0 + y_0 i$, both \left\{ \begin{aligned} \frac{\partial u}{\partial x} &= \frac{\partial v}{\partial y} \\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{aligned} \right. must be satisfied at $z_0$. Let's check the first equation: \begin{align} \frac{\partial u}{\partial x} = -\frac{2x}{(x^2+y^2)^2} \overset{?}{=} -\frac{2y}{(x^2+y^2)^2} = \frac{\partial v}{\partial y} \end{align} We can cancel all the common factors to obtain $$x = y$$ So the first of the CR equations holds whenever $x = y$. As for the second equation:$\newcommand{\AND}{\ \ {\rm{\small{AND}}}\ \ }$ $$\frac{\partial u}{\partial y} = -\frac{2y}{(x^2+y^2)^2} \overset{?}{=} \frac{2x}{(x^2+y^2)^2} = -\frac{\partial v}{\partial x}$$ This simplifies to $$-y = x$$ So both CR equations are satisfied only when $x = y \AND -y = x$. This only happens when $x = y = 0$, which is already precluded because $f$ is not defined there. So the CR equations are satisfied nowhere, and hence $f$ is differentiable nowhere, and is therefore holomorphic nowhere.

Let the operators $D_1$ and $D_2$ be defined as $$D_1(u+iv) = u_x - v_y \ \text{ and } \ D_2(u+iv) = u_y + v_x.$$ Then $u+iv$ is analytic at $z_0 = x_0 + i y_0$ if and only if $u(x_0,y_0)+iv(x_0,y_0)$ belongs to $\ker D_1 \cap \ker D_2$.