Introduction to the Residue Theorem
The residue theorem is a crucial result in complex analysis that connects the contour integral of a complex function to the sum of its residues at singular points enclosed by the contour.
Statement of the Residue Theorem
- Consider a complex function (f(z)) with singular points (z_1, z_2, ..., z_n).
- The integral of (f(z)) around a closed contour (C) enclosing these points is:
[\oint_C f(z),dz = 2\pi i \sum_{j=1}^n \text{Res}(f, z_j)]
where (\text{Res}(f, z_j)) denotes the residue of (f) at (z_j).
Proof Overview and Diagrammatic Approach
- Draw a large closed contour enclosing singular points (z_1, z_2, z_3) for simplicity.
- Create small circles (P_1, P_2, P_3) around each singular point and introduce cuts to define a new contour (C_1) that contains no singularities.
- By Cauchy's theorem, the integral over (C_1) is zero.
- This relates the integral over (C) to the integrals over the smaller circles around singularities.
Expansion Using Laurent Series
- Around each singular point (z_j), expand (f(z)) as a Laurent series:
[f(z) =\sum_{n=-\infty}^\infty b_{n}^{(j)} (z - z_j)^n] \ - The contour integral around the small circle (P_j) splits into parts over analytic and principal parts.
- Integrals over analytic parts vanish due to holomorphicity and Cauchy-Riemann relations.
Evaluating the Integral of the Principal Part
- Parametrize the small circle (P_j) as (z = z_j + \rho_j e^{i\theta}, \theta \in [0,2\pi]).
- Calculate integrals for terms (b_n^{(j)} (z - z_j)^n) where (n<0).
- For all terms except when (n = -1), the integral evaluates to zero.
- The (n = -1) term gives the nonzero contribution defining the residue.
Final Result and Interpretation
- The integral around each singularity (z_j) is (2\pi i) times its residue (b_{-1}^{(j)}).
- Summing over all singularities inside (C) yields the residue theorem's formula.
- If no singularities are enclosed, the integral is zero, aligning with Cauchy's theorem.
Key Takeaways
- The residue corresponds to the coefficient of ((z - z_j)^{-1}) in the Laurent series.
- Contour integrals in complex analysis depend solely on residues at enclosed singularities.
- The theorem bridges local behavior near singularities with global integral properties.
Related Concepts
- Connection to Cauchy's integral formula and theorem.
- Importance for evaluating complex integrals and applications in physics and engineering.
Understanding the residue theorem through this proof enriches comprehension of complex function behavior and integral evaluation techniques essential for advanced mathematical and physical applications. For a deeper grasp on Laurent expansions and residue calculations, see Understanding Laurent Series and Residues in Complex Analysis.
Welcome back to my guide on complex variables. Today's video is going to be about the residue theorem. This is
probably one of the most important theorems in complex variables. Here's how it works. Say I have a complex
function f of z with a bunch of singular points, say z, z1, z2 all the way to zn. What the residue theorem says is that
the contour integral of f of z over a curve which encloses all of these singular points is just 2 pi i times the
sum of the residues of f at each of those singular points. Just for your interest, I'm going to prove this
theorem to you. Let's start by making a diagram to help us. Say we have a big closed contour in the complex plane that
encloses these three singular points inside it. Z1, z2, and z3. There could be more but for now I'm just drawing
three for the sake of simplicity. The contour integral of f over this entire region is only due to the contributions
of these singular points. Why? Well, if I draw three really small circles over these singular points and make a bunch
of small cuts at these circles, then since this resulting contour, I'll call it C1, doesn't enclose any singular
points, its contour integral is zero, thanks to Koshi's theorem, which we covered a couple of videos ago. If I
made the size of these gaps approach zero, then I would find that the anticlockwise integral around this
purple curve C plus the clockwise integrals around each of these circles P1, P2, and P3 equals the anticlockwise
integral along the entire curve C1. It's very similar to what we did last time. Since we know that the ladder is zero,
we're just left with this expression. I can now isolate the integral around the entire curve C and use the fact that
the clockwise integrals are just the negatives of the anticlockwise integrals to show that the integral of a complex
function over a contour C is only due to the contributions of the singular points inside that contour. And this is where
the LA series comes in. Around each of these singular points Z1, Z2, and Z3, which correspond to the enclosing
circles P1, P2, and P3, I can expand f of z as a la series. Let's look at the la series of f around z1. Note the
presence of the extra one in the subscript to denote the fact that these are the raw series coefficients around
z1. If I take the integral of f over the contour p1, then I know I can split that integral up into two parts. One of which
is the integral over the analytic part of the lauron series and the other is an integral over the principal part. I can
switch the order of the integration and summation which allows me to rewrite this expression as the following. Now
the integrals for the analytic part of the Lauron series all disappear because of Cowshi's theorem since the contour
integral of a function over a region where it's holorphic is zero and this analytic portion corresponds to the part
of the function that is holorphic over that region. So now we're left with the
integrals over just the principal part of the Lauron series. If we look at this last integral, the one which I have b
subj1 starting from the index of two, then what I can do is that because p1 is just a circle say of radius row 1, I can
write my complex variable z as z1 + row 1 e to the i theta which is just a polar representation of a circle in the
complex plane. If I take the differential of this then I can see that dz is just row 1 * i * e i theta d theta
where theta varies from 0 to 2 pi. Now when I'm doing a contour integral around the whole circle that's the same as
keeping the radial position row 1 constant and varying theta from 0 to 2 pi. Because of this, I can write my
integral as the integral from 0 to 2 pi of b subj1 / row 1 * e i theta to the power j * row 1 * i * e i theta d theta.
I can simplify this expression to leave myself with only a single exponential. So I get b subj1 * i / row 1 ^ j -1 *
the integral from 0 to 2<unk> of the exponential of i * j -1 * theta d theta. Integrating this gives us b subj1 * i
row 1 j1 * -1 / i * jus1* the exponential of i * j -1 * theta. The limits are from 0 to 2 pi.
Now if I apply the limits then I can see that at theta= 0 my exponential is just the exponential of 0 which is 1. And at
theta= 2 pi I can apply oilers's formula which I'll write over here for your own verification. By Oilers's formula I know
that the sign term is zero in both cases since j is an integer and the sign of an integer multiple of 2 pi is always zero.
I also know that the cosine of 2 pi and the cosine of 0 are both 1. So my integral expression would just be 1
minus 1 which is zero. Note that this is true only for integer values of J that aren't one. If J did turn out to be one,
then this denominator would be undefined and we wouldn't be able to use this integral. So if we go back to the
integral over the circles P1 of the complex function f of Z, then the only term we're left with is the B1 term of
the Laur series or the residue term. In fact, we can easily find this integral by applying Koshi's integral formula
that we touched on in the last video. What we end up with is that the closed integral in the anticlockwise direction
along P1 of F of Z DZ is just 2 pi * B11, where B11 is the residue of the function expanded around the point Z1.
Now, you can see why the coefficient B11 is called a residue because it's the only term that's left when we do the
contour integration of the Lauras series. Now that I've shown this result for one singular point inside the closed
contour C, I can show the exact same thing for the other singular points Z2 and Z3. The end result is that when
there are three singular points, the contour integral of f of z over the closed curve c is just 2 pi i * the sum
of the residues of the lauron expansions at z1, z2, and z3. If there is more than three singular points, this naturally
leads to the more general result of the residue theorem, which is that the contour integral of f of z over a curve
which encloses a bunch of singular points is 2<unk>i i * the sum of the residues of f at each of those singular
points. A couple of things to note before I go. One is that if there are no singular points enclosed by C, then the
integral is zero because the residue of a non-s singular point is zero since the entire principal part of the Lauron
series around the non-s singular point would of course be zero. Another thing to note is that particularly at the
beginning, the proof of the residue theorem started off very similar to the proof of the Koshi's integral formula.
So if you understood it last time, hopefully you understood it this time and vice versa. If you still don't get
it, you can always write an angry comment below and I'll be sure to respond to you passive aggressively.
Anyway, that does it for the residue theorem. Thanks for watching.
The residue theorem states that the contour integral of a complex function around a closed path equals 2πi times the sum of the residues at singularities inside the contour. It simplifies evaluating complex integrals by linking local behavior near singular points to global integral values, making it essential for solving problems in mathematics and physics.
To find the residue at a singularity z_j, expand the function into a Laurent series around z_j and identify the coefficient b_{-1} of the (z - z_j)^{-1} term. This coefficient is the residue, which directly determines the integral's contribution from that singularity.
Analytic parts correspond to non-negative powers of (z - z_j) and represent holomorphic functions within the contour. Their integrals over closed paths vanish due to Cauchy's theorem and Cauchy-Riemann relations, ensuring only the principal part, specifically the (z - z_j)^{-1} term, contributes to the integral.
Yes, if no singularities lie inside the contour, the residue theorem implies the contour integral is zero, aligning with Cauchy's theorem. This means holomorphic functions inside the contour have zero integral around the path, simplifying integral evaluations in such cases.
Parametrizing small circles P_j around each singularity as z = z_j + ρ_j e^{iθ} allows explicit evaluation of contour integrals term-by-term in the Laurent series. This approach reveals that only the (z - z_j)^{-1} term yields a nonzero integral, defining the residue and validating the residue theorem.
The residue theorem generalizes Cauchy's integral formula by connecting contour integrals to sums of residues at singularities inside the contour, rather than just evaluating a function's value at a point. Both rely on holomorphic function properties, but the residue theorem extends integral evaluation to functions with singularities.
In physics and engineering, the residue theorem is instrumental for calculating complex integrals that arise in signal processing, fluid dynamics, electromagnetism, and quantum mechanics. By evaluating residues at singularities, practitioners can solve integral equations and analyze system behaviors efficiently.
Heads up!
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