12-13: Complex analysis introduction & Cauchy-Riemann equations Flashcards
What is the definition of an analytic function?
A complex function f(z) is said to be analytic at the point z_0 if the derivative at this point is unique. Furthermore, f(z) is analytic within a region or domain if at each point in the domain, the derivative is defined and unique.
What is the point z_0 called if a function f(z) is not analytic at a point z_0?
The point z_0 is called a singularity or a pole.
What is it called if a function f(z) is both single valued and analytic at z_0?
The function f(z) is called regular at z_0
What are the Cauchy-Riemann equations?
The Cauchy-Riemann equations provide a simple necessary and sufficient test of whether a complex function is analytic
Derive the Cauchy-Riemann equations
- Start with complex function f(z) = u(x, y) + iv(x, y), where u and v are real functions
- Assume function is analytic in a region D, then take the derivative using lim_{δz→0} = {f(z + δz) − f(z)}/δz, seperating into complex and real parts
- Take δy -> 0 then δx -> 0
- Repeat for δx -> 0 and δy -> 0
- Set results of two methods equal
- Equate real and imaginary parts to get Cauchy-Riemann equations
We have proven that if is analytic in a region D, then the above equations hold for all z inside D.
If Cauchy-Riemann equations prove that f(z) is analytic in a region D, what does this imply for all z inside D?
That it holds for all z inside D. All of z is analytic inside D.
If a pair of real-valued functions u(x,y) and v(x,y) satisfy the above two partial differential equations within a domain D, where z is analytic, what does this imply about the function?
The converse is also true: If a pair of real-valued functions u(x,y) and v(x,y) satisfy the above two partial differential equations within a domain D, the complex function f(z) is analytic in D where:
f(z) = u(x, y) + iv(x, y)
What is a single valued function?
It means that every element maps to exactly one value. Because in complex analysis you can sometimes dealwith things like x^{n/2}, which has n solutions if x
is non-zero.
For for what case are the cauchy equation derivatives not defined? What does this mean for the analyticness of the function?
Derivatives are not defined at x=y=0 (i.e. z=0). So function is analytic everywhere EXCEPT z=0.
When can we safely derive f(z) in a normal way as you would for real functions, ie df/dz?
After using the Cauchy-Riemann equations to figure out the function f(z) is analytic and this function will be unique.
Derive that d^2u/dx^2 + d^2u/dy^2 = 0 if f(z)=u+iv is analytic. This takes of form of what other equation in 2D?
- Take derivative twice of du/dx. Apply CR equation to swap du/dx for dv/dy
- Take derivative twice of du/dy. Apply CR equation to swap du/dy for dv/dx
- Add resultant equations from (1) and (2)
End up with Laplace’s equation in 2-D so u (or v) solve for this. Works both ways: any function u(x,y) satisfying Laplace’s equation in a region is real (or imaginary) part of analytic function f(z). [can work out f(z) simply from knowing u and using the C-R equations]
