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#1 2013-05-15 21:23:27

Ivar Sand
Registered: 2013-01-22
Posts: 14

A proof of the residue theorem (complex analysis)

In short. Polar coordinates are used to prove the residue theorem for functions represented by a single Laurent series. The path of integration may have a general shape. The reader should be familiar with complex analysis.

Background. The proof emerged from the idea that the proof should contain terms with no contribution to the integral corresponding to the fact that there is no "height difference between the starting and ending points when walking around a mountain top and ending at the starting point". This, and a wish to use as little as possible of the theory of complex analysis, lead to the use of Laurent series and the introduction of a polar coordinate representation in the complex plane.

Complex analysis is the study of complex functions mapping the complex space ℂ to ℂ. A number z in ℂ has two components, a real component and an imaginary component, which is a real number multiplied by i where i=√-1. Nevertheless, algebraically, we regard z as we do a number in ℝ, i.e. as a single entity.

An analytic function is a function that can be represented by a Taylor series. A meromorphic function is a function that is analytic except at isolated points where it has poles. Locally around a point in ℂ, a meromorphic function can be represented by a Laurent series about that point. A Laurent series is a Taylor series to which terms with negative exponents are added. To represent a meromorphic function throughout its domain, several Laurent series may be required.

An integral of a complex function is like a line integral of a scalar real function mapping ℝ² into ℝ.

Laurent series. We identify a meromorphic function f(z) by its Laurent series about a point c,


where z ∈ D, an open disc with radius > 0 and centre at c. f is analytic everywhere on D except at c where f may have a pole[sup]1[/sup].

A Laurent series can be viewed as the sum of two series. The part with index n ranging from -∞ to -1 converges on ℂ - {c}, and the other part with n ranging from 0 to ∞ converges on D. These two series are power series, in 1/(z-c) and z-c respectively.

A Laurent series is integrable term by term along rectifiable paths within closed and bounded subsets of D - {c}.[sup]2[/sup]

Parameterised integration path. Let P be a rectifiable (i.e. P has a finite length)[sup]3[/sup], closed, and connected path, which is contained in a closed and bounded subset of D - {c}. Let a parameterisation of P be given,
    P: z=z(t), t∈[α,β]⊂ℝ, α<β,
    the angular part of z(t)-c increases by 2π as t varies from α to β,
    z(t) is continuous,
    z(t) is piecewise continuously differentiable[sup]4[/sup] on [α,β].
The equation
    z(t) = r(t)e[sup]iθ(t)[/sup] + c
defines the radial function r(t) = |z(t)-c| and the angular function[sup]5[/sup] θ(t) = arg(z(t)-c). The properties of r(t) and θ(t) correspond to the ones of z(t) (stated without proof) and are,
    r(t) is continuous,
    r(t) is piecewise continuously differentiable[sup]4[/sup] on [α,β]
    θ(t) is continuous,
    θ(t) is piecewise continuously differentiable[sup]4[/sup] on [α,β].

Polar coordinates in ℂ. We calculate the derivative of z(t)-c, which equals z'(t),




r(t) in the denominator above is never 0 on P because P does not contain c.
We use the last expression for z'(t) above in the first of the following definitions,
    dz = z'(t)dt
    dr = r'(t)dt
    dθ = θ'(t)dt
and get,
    dz = z'(t)dt
    = r'(t)(z(t)-c)/r(t)dt + i(z(t)-c)θ'(t)dt
    = (z(t)-c)/r(t)r'(t)dt + i(z(t)-c)θ'(t)dt
    = (z-c)/r dr            + i(z-c)dθ
from which we identify the radial part, d[sub]r[/sub]z, and angular part, d[sub]a[/sub]z, of dz,

    dz = d[sub]r[/sub]z + d[sub]a[/sub]z
    d[sub]r[/sub]z = (z-c)/r dr
    d[sub]a[/sub]z = i(z-c)dθ.
Note in the expressions of the differentials of the complex polar coordinates, d[sub]r[/sub]z and d[sub]a[/sub]z, the corresponding differentials of the real polar coordinates, dr and dθ.

The residue theorem for a function represented by a single Laurent series.


The direction of integration is counterclockwise around the point c, see the description of P above.
Proof. Since P is a rectifiable path in a closed and bounded subset of D - {c}, we can integrate the Laurent series term by term,


Thus we are led to consider ∮[sub]P[/sub](z-c)[sup]n[/sup] dz, n= …, -2, -1, 0, 1, 2, …, and we need to prove,


where δ[sub]n,-1[/sub] = 1 if n = -1 otherwise 0.

Now we introduce polar coordinates in the left-hand side of (1),





We separate the n=-1 part from the rest,



We use the fact that n=-1 in the n=-1 part. Also, in the third integral, we make use of the parameterisation of P,



(Note that since r'(t) is only piecewise continuous on [α,β], the integrand of the integral above over [α,β] may not always be defined. This problem is circumvented by defining the value of this integral as the sum of the integrals with the same integrands as the integral above but with integration intervals that are the subintervals of [α,β] where r'(t) is continuous.[sup]4[/sup])
We observe ∮[sub]P[/sub] 1/r dr=0; this is an example of the "walking around a mountain" effect we talked about. Also, since we integrate once around c in a counterclockwise direction, ∮[sub]P[/sub]dθ=2π,



We observe that we have already gotten the result we are looking for in line one, so we aim at proving that the expression within the last parenthesis is zero,



r[sup]n[/sup] is the derivative of r[sup](n+1)[/sup]/(n+1),



Since r(t) and θ(t) are continuous and piecewise continuously differentiable on [α,β], so are r(t)^(n+1)/(n+1) and e^iθ(t)(n+1). This allows us to use integration by parts for the integral over [α,β], see Lemma 1 below,








Now we use r(β)=r(α) and θ(β)=θ(α)+2π,









e[sup]i2π(n+1)[/sup]=1 yields,





Two elements have opposite signs and cancel each other; they are another example of the "walking around a mountain" effect. We are left with only two integrals within the last parenthesis,





We find that these two integrals cancel each other,













This is the right-hand side of (1), and the proof is complete.

Lemma 1 (integration by parts).


where g and h are complex-valued functions.
Proof. We separate the real and the imaginary parts, g(t) = g[sub]1[/sub](t) + ig[sub]2[/sub](t) and h(t) = h[sub]1[/sub](t) + ih[sub]2[/sub](t) where g[sub]1[/sub](t), g[sub]2[/sub](t), h[sub]1[/sub](t), and h[sub]2[/sub](t) are real-valued functions.

We substitute the real-valued functions for the complex-valued functions on the left-hand side of the equation in the lemma,







The functions in the integrands of these four integrals are real-valued functions, and that allows us to use integration by parts,[sup]6[/sup]













This expression is identical to the right hand side of the equation in the lemma, and the proof is complete.

1. Laurent series
2. Laurent series – Uniqueness
3. Curve – Lengths of curves
4. Reinhold Remmert, Theory of Complex Functions, p. 173-174.
5. Argument (complex analysis)
6. Integration by parts

Last edited by Ivar Sand (2013-07-10 21:20:59)

I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway.


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