- #1
Bashyboy
- 1,421
- 5
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?
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?