Find Analytic Function F: Cauchy-Riemann Eqns

[morganmkm]

find an analytic function F where F' = f

f(z) = z-2
f(z) = ((z^4) +1)/(z^2)
f(z) = sinzcosz


I know I have to put these into the cauchy riemann equations but I don't know what to use for my du/dx or du/dy because I am not sure to use x-2 for my du/dx etc or if I am supposed to integrate first. I don't know which parts of the equations fit into the riemann-cauchy equations. My book only gives one example and I don't know how to relate these, please help

 

[micromass]

Just integrate like in calculus?

 

[mathman]

All these expressions can be handled by ordinary integration (the second one is more difficult).
For example f(z) = z-2, them F(z) = z²/2 - 2z + C.

 

[Bacle2]

The second function is not entire , so you need to specify a domain. Depending on the level of your class, it would be nice to have an argument for why the antiderivative is analytic (hint, use something like independence of path and Morera's thm.)

 

[blue_raver22]

To find an analytic function F, we need to satisfy the Cauchy-Riemann equations, which are:

∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x

where u and v are the real and imaginary parts of F, respectively. In order to find F, we need to integrate the given functions for f(z) and then use the Cauchy-Riemann equations to determine the real and imaginary parts of F.

For the first function, f(z) = z-2, we can integrate to get F(z) = (z^2)/2 - 2z + C, where C is a constant. Using the Cauchy-Riemann equations, we can see that ∂u/∂x = z - 2 and ∂v/∂y = 0. Therefore, we can set u = (z^2)/2 - 2z and v = 0, and F(z) = (z^2)/2 - 2z is an analytic function with f(z) = z-2.

For the second function, f(z) = ((z^4) +1)/(z^2), we can integrate to get F(z) = (z^3)/3 + z + C. Using the Cauchy-Riemann equations, we can see that ∂u/∂x = (z^3)/3 + 1 and ∂v/∂y = 0. Therefore, we can set u = (z^3)/3 + 1 and v = 0, and F(z) = (z^3)/3 + 1 is an analytic function with f(z) = ((z^4) +1)/(z^2).

For the third function, f(z) = sinzcosz, we can use the identities sinz = (e^(iz) - e^(-iz))/2i and cosz = (e^(iz) + e^(-iz))/2 to rewrite f(z) as f(z) = (e^(2iz) - 1)/4i. We can then integrate to get F(z) = -(e^(2iz))/8i - z/4 + C. Using the Cauchy-Riemann equations, we can see that ∂u/∂