Cauchy rieman equation. What is u?

Question

Using cauchy rieman equation, i want to show the function is analytic. So i want to decompose from f(z) to two term (real part and imaginary part) With rectangular form or polar form. But it is so long working.. is there efficient way?

Answer 1

Certainly: a function $f(z,\bar{z})$ is analytic iff

$$\frac{\partial f}{\partial \bar{z}} = 0$$

For example, $f(z,\bar{z}) = \arg{z} = -\frac{i}{2} [\log{z} - \log{\bar{z}}]$ is not analytic. However, the function you point out, $f(z,\bar{z}) = 3 \pi^2/(z (z^2+4 \pi^2)) $ is most certainly analytic because of the lack of dependence on $\bar{z}$.

However, if you insist on breaking up the function into real and imaginary parts, it helps to use partial fractions. In this case:

$$f(z,\bar{z}) = \frac1z - \frac12 \left (\frac1{z-i 2 \pi} + \frac1{z+i 2 \pi} \right ) $$

or, letting $z=x+i y$ and $f=u + i v$, we have

$$u+i v = \frac{x-i y}{x^2+y^2} - \frac12 \left [\frac{x-i (y-2 \pi)}{x^2+(y-2 \pi)^2} + \frac{x-i (y+2 \pi)}{x^2+(y+2 \pi)^2} \right ] $$

Then

$$u(x,y) = \frac{x}{x^2+y^2} - \frac{x}{2} \left [\frac{1}{x^2+(y-2 \pi)^2} + \frac{1}{x^2+(y+2 \pi)^2} \right ] $$

$$v(x,y) = -\frac{y}{x^2+y^2} + \frac{1}{2} \left [\frac{y-2 \pi}{x^2+(y-2 \pi)^2} + \frac{y+2 \pi}{x^2+(y+2 \pi)^2} \right ] $$