13-14: Contour integration, Laurent Series & Residue theorem Flashcards
State Cauchy’s first integral theorem.
If a function f(z) is analytic over a region D, any integral around a closed loop in that region is 0. ie the integral only depends on the endpoints.
Derive Cauchy’s first integral theorem from a function f(z) which is analytic with the region D and is simply connected (has no holes and has singularities / is well-defined )
Integrate f clockwise or counter-clockwise over some closed path C that is within D
1. Seperate (u + iv)(dx + idy) into real and imaginary parts
2. Apply green’s theorem.
3. Note that derivatives inside double integrals now take form of Cauchy-Riemann equations, which can be applied since function is analytic over the region D that the loop sits inside.
4. therefore, closed loop integral of f(z)dx=0
We can conclude that the integral of f(z) around ANY closed loop in D is zero. Therefore the integral between any two points is independent of the path joining them.
State Green’s theorem and describe how it relates to Stoke’s theorem.
[in page 5 of handwritten self notes, or in lecture slides].
It is stokes theorem in 2D
What is a common name of the singularity that we expand the Laurent series about?
A pole. (point of the function that is not analytic)
What do these negative terms in the Laurent series sometimes have?
A finite number of terms like (z-z_0)^{-n}
If the highest negative power in the series is m, what do we say about the Laurent series?
If the highest negative power in the series is m, then we say the singularity is a pole of
order m. If a_m ≠ 0, but all the a’s after a_m are zero, f(z) is said to have a pole of order m at z=z_0.
If in a Laurent Series, the highest order pole is order 1, what do we call this?
If the highest order pole is order 1, then this is called a simple pole.
What do we call it if the singularity in the Laurent Series has infinitely many terms with negative power?
The singularity is called an essential singularity at z=z_0.
In the Laurent Series, if f(z) is analytic everywhere, what happens to f(z)? How do you get to this stage if you have a smaller circle (C_1 with radius R_1) inside another larger circle (C_2 with radius R_2)?
It becomes a normal Taylor series and can be expanded about any point in which f(z) is analytic.
To get to this stage, shrink inner circle C_1 to have 0 radius (integral equation in notes). Hence R1 becomes zero. a_{-n}=0. This makes the laurent series a normal taylor series with positive powers to n, exact same to that of a regular function. The domain of the series is just the circle C_2.
What name do we give to the part of the Laurent series with negative powers?
The principle part.
What is the residue of a pole in the Laurent series?
It is the coefficient a_{-1}, i.e. the coefficient in front of 1/(z-z_0) in the Laurent series.
How can you prove the Residue theorem supposing z is analytic except for poles at a, b and c over a closed loop C?
- Shrink C to make circles around each pole.
- Lines are equal and opposite (diagram in pg 12 of slides) thus cancel out and are left with small circles around each pole to integrate over.
- Expand f(z) as a Laurent series around each pole.
- ∳_0 1/z^p = 0 (for p>1), ∳_0 dz/z = 2πi and terms with non-negative powers are zero by Cauchy’s theorem (z^p is analytic for p ≥ 0
- Result is ∳a f(z)dz = 2πia{-1}. So for all total closed loop ∳C
f(z)dz = 2πi(a{−1} + b_{−1} + c_{−1})
Integral of f(z) around the contour C is equal to 2πi times the sum of the residues at the poles enclosed by C.
What does the Residue Theorem state?
That ∳_C f(z)dz = 2πi∑ residues at poles of f(z) within C
Evaluate integral of form
∫^∞_{−∞} P(x)/Q(x) dx
Answer in page 708 of textbook and q. 16 in QS 6
