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

  1. Oct 28, 2014 #1
    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. jcsd
  3. Oct 28, 2014 #2
    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.
     
  4. Oct 28, 2014 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

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