# Determining At Which Points A Function is Analytic (Holomorphic) and Singular

1. Oct 28, 2014

### Bashyboy

Hello everyone,

I have to determine at which points the function $\displaystyle f(z) = \frac{1}{z}$ is analytic.

I just want to verify that I understand what the Cauchy-Riemann equations tells us about a function, in terms of its differentiability, and what it means for a function to be analytic (holomorphic).

Let's look at the first function.

$\displaystyle f(z) = \frac{1}{z} = z^{-1} = \frac{x}{x^ + y^2} + i \frac{-y}{x^2 + y^2}$. This gives us the real function

$\displaystyle u(x,y) = \frac{x}{x^2 + y^2}$, which has the partials

$\displaystyle u_x = \frac{x \cdot 2x - 1 \cdot (x^2 + y^2)}{(x^2 + y^2)^2} = \frac{2x^2 - x^2 - y^2}{(x^2+y^2)^2} = \frac{x^2 - y^2}{(x^2+y^2)^2}$

and

$\displaystyle u_y = \frac{x \cdot 2y - 0 \cdot (x^2 + y^2)}{(x^2 + y^2)^2} = \frac{2xy}{(x^2+y^2)^2}$

The imaginary function is $\displaystyle v(x,y) = \frac{-y}{x^2 + y^2}$, which has the partials

$\displaystyle v_x = -y \cdot (x^2 + y^2 )^{-2} (2x) = \frac{-2xy}{(x^2 + y^2)^2}$

and

$\displaystyle v_y = \frac{-y \cdot (2y) - (-1) \cdot (x^2 + y^2)}{(x^2 + y^2)^2} = \frac{-2y^2 + x^2 + y^2}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2+y^2)^2}$

Now, there is a theorem which states that a function $f(z)$ is differentiable at a point if that point satisfies the conditions

$\displaystyle u_ x = v_y$ and $\displaystyle u_y = - v_x$ (Cauchy-Riemann equations)

Consequently, any point $z$ which does not satisfy these equations is a point at which $f(z)$ is not differentiable.

Clearly, if I substitute in the partials, I will find that the equations are satisfied for every complex number, except $z=(0,0)$. Therefore, the function is differentiable everywhere, except $z=(0,0)$.

Now, I claim that the function is analytic everywhere because I can form an open disk around around any point in the plane, and the disk will only enclose points at which $f(z)$ is differentiable.

Does this sound correct?

2. Oct 28, 2014

### Bashyboy

Actually, I am not certain that it would be analytic at $z=(0,0)$. Here are the definitions of analyticity in an open set and at a given point $z_0$:

1) A function of a complex variable $z$ is analytic (regular, holomorphic) in an open set if
it has a derivative at each point in the set. (Note that here derivatives and analyticity
are not defined on boundaries - only in neighborhoods.)

2) A function of a complex variable $z$ is analytic at a point $z_0$ if it is analytic in some
open set containing $z_0$.

If I understand these correctly, which I previously thought I did, then for $f(z)$ to be analytic at the point $z_0$, I must be able to find an open disk which includes $z_0$ and the function must be differentiable at every point in the disk--which would include $z_0$. If this is so, then $z=0$ would not be analytic. Consequently, it would be a singular point.

3. Oct 28, 2014

### Dick

Right. It's analytic at every point except z=0.