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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. 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^{1}. 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/(zc) and zc respectively. A Laurent series is integrable term by term along rectifiable paths within closed and bounded subsets of D  {c}.^{2} Parameterised integration path. Let P be a rectifiable (i.e. P has a finite length)^{3}, 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, z(β)=z(α), 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^{4} on [α,β]. The equation z(t) = r(t)e^{iθ(t)} + c defines the radial function r(t) = z(t)c and the angular function^{5} 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(β)=r(α), r(t) is continuous, r(t) is piecewise continuously differentiable^{4} on [α,β] and θ(β)=θ(α)+2π, θ(t) is continuous, θ(t) is piecewise continuously differentiable^{4} 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 = (zc)/r dr + i(zc)dθ from which we identify the radial part, d_{r}z, and angular part, d_{a}z, of dz, dz = d_{r}z + d_{a}z d_{r}z = (zc)/r dr d_{a}z = i(zc)dθ. Note in the expressions of the differentials of the complex polar coordinates, d_{r}z and d_{a}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 ∮_{P}(zc)^{n} dz, n= …, 2, 1, 0, 1, 2, …, and we need to prove, where δ_{n,1} = 1 if n = 1 otherwise 0. Now we introduce polar coordinates in the lefthand 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.^{4}) We observe ∮_{P} 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, ∮_{P}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^{n} is the derivative of r^{(n+1)}/(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^{i2π(n+1)}=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 righthand side of (1), and the proof is complete. Appendix. Lemma 1 (integration by parts). where g and h are complexvalued functions. Proof. We separate the real and the imaginary parts, g(t) = g_{1}(t) + ig_{2}(t) and h(t) = h_{1}(t) + ih_{2}(t) where g_{1}(t), g_{2}(t), h_{1}(t), and h_{2}(t) are realvalued functions. We substitute the realvalued functions for the complexvalued functions on the lefthand side of the equation in the lemma, The functions in the integrands of these four integrals are realvalued functions, and that allows us to use integration by parts,^{6} This expression is identical to the right hand side of the equation in the lemma, and the proof is complete. References. 1. Laurent series http://www.encyclopediaofmath.org/index.php/Laurent_series 2. Laurent series – Uniqueness http://en.wikipedia.org/wiki/Laurent_series#Uniqueness 3. Curve – Lengths of curves http://en.wikipedia.org/wiki/Rectifiable_path#Lengths_of_curves 4. Reinhold Remmert, Theory of Complex Functions http://books.google.no/books?id=uP8SF4jf7GEC&pg=PA173&lpg=PA173&dq=integrate+piecewise+continuously+differentiable&source=bl&ots=yMrRr6wE4v&sig=izem7hy8bHrw8TkKSNEvhujMrs&hl=no&sa=X&ei=dpPWUbSzGoOv4QS_poC4DA&redir_esc=y#v=onepage&q=integrate%20piecewise%20continuously%20differentiable&f=false, p. 173174. 5. Argument (complex analysis) http://en.wikipedia.org/wiki/Argument_(complex_analysis) 6. Integration by parts http://en.wikipedia.org/wiki/Integration_by_parts Last edited by Ivar Sand (20130711 19:20:59) I majored in Physics in 1976. Also, I studied mathematics and computer science. I work as a computer programmer. I am from Norway. 