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Calculus and Beyond Homework Help
Complex valued line integrals
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[QUOTE="Mathman23, post: 1313495, member: 14024"] Hi there I got a couple of question regarding the topic above [h2]Homework Statement [/h2] (a) Given the integrals [tex]\int \limit_{0}^{i} \frac{dz}{(1-z)^2}[/tex] [tex]\int_{i}^{2i} (cos(z)) dz [/tex] [tex]\int_{0}^{i\pi} e^{z} dz[/tex] (1)write this as a Line integral on the form [tex]\int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt[/tex] and (2)Next find sum the of integrals using anti-derivatives. (b) Here am I unsure. How do I approach to calculate ? [tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex] where is [tex]\gamma_{n}:[0,2\pi] \rightarrow \mathbb{C} [/tex] is a parameter presentation of the unit circle, where [tex]n \in \mathbb{Z} - \{0\}[/tex] and which runs through [tex]\gamma_{n}(t) = e^{itn}[/tex] [h2]Homework Equations[/h2] Have I understood and solved (a) correctly? What about (b) could somebody please be so kind give me a hint/(some help) :) ? [h2]The Attempt at a Solution[/h2] attempted Solution A: I choose [tex]\gamma(t) = t \cdot i[/tex] where [tex]t \in [0,1][/tex] Since [tex]\gamma(t)' = i[/tex] then the solution is [tex]\int_{\gamma} \frac{dt \cdot i}{(1-(it))^2} = \int_{0}^{1} \frac{dt \cdot i}{(1-(it))^2} = -1/2 + 1/2 \cdot i[/tex] the finding the sum of the original integral [tex]\int_{0}^{i} \frac{dz}{(1-(z))^2} = -1/2 + 1/2 \cdot i [/tex] Second integral: [tex]\int_{0}^{i\pi} e^{z} dz = \int_{\gamma} (e^{it} \cdot i) dt = \int_{0}^{\pi} (e^{it} \cdot i) dt = -2 [/tex] finding the sum of the integral: [tex]\int_{0}^{i\pi} e^{z} dz = -2[/tex] (B) Attempted solution If [tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex], then to solve this integral using the definition of the line integral then I take [tex]\int_{\gamma_{n}} \frac{dz}{z} = \int_{0}^{2\pi} \frac{1}{e^{t \cdot n \cdot i}} \cdot \frac{d}{dt}(e^{t \cdot n \cdot i}) dt = 2 \cdot n \cdot \pi \cdot i[/tex] where [tex]n \in \mathbb{Z} - \{0\}[/tex] Could this be the solution? Best Regards Fred [/QUOTE]
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Complex valued line integrals
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