Evaluating Complex Contour Integrals Using the Residue Theorem
Contour integrals play a vital role in complex analysis, serving as a powerful tool for evaluating integrals within a specified domain. This article delves into the process of solving a complex contour integral problem using the Residue Theorem. The focus will be on the methodology, encountered challenges, and the underlying mathematical principles.
Introduction to Contour Integrals
Contour integrals are integral operators that operate over a contour, typically a closed curve in the complex plane. The context of this article is to evaluate the contour integral given by:
[ int_{z1} frac{e^{frac{z}{sqrt{3} - 2}}}{z^2 4iz 1} dz ]
Identifying Poles and Applying the Residue Theorem
The first step in evaluating such an integral is to identify the poles of the integrand within the specified contour. A pole occurs where the denominator of the integrand equals zero, while the numerator remains non-zero.
The poles can be determined by solving the equation:
[ z^2 4iz 1 0 ]
Solving for ( z ), we get:
[ z -2 pm isqrt{5} ]
Inspecting these solutions, only one lies within the contour of integration. Specifically, the pole at ( z -2 isqrt{5} ) is the only one contained within the unit circle ( z 1 ).
Applying the Residue Theorem
The Residue Theorem states that the value of a contour integral around a closed path is equal to ( 2pi i ) times the sum of the residues of the function at each of the poles inside the closed path. Here, we apply this theorem to the integral.
Rescaling the factor in the denominator:
[ frac{e^{frac{z}{sqrt{3} - 2}}}{z^2 4iz 1} frac{e^{zsqrt{3} - 2}}{z - (-2 isqrt{5}) (z - (-2 - isqrt{5}))} ]
We simplify the integral as follows:
[ int_{z1} frac{e^{frac{z}{sqrt{3} - 2}}}{z^2 4iz 1} dz int_{z1} frac{frac{e^{zsqrt{3} - 2}}{z 2isqrt{5}}}{z - (-2 isqrt{5})} dz ]
Using the Residue Theorem, we find that:
[ 2pi i cdot frac{e^{frac{z}{sqrt{3} - 2}}}{z 2isqrt{5}} Bigg|_{z -2 isqrt{5}} ]
Simplifying this expression, we obtain:
[ 2pi i cdot frac{e^{sqrt{3} - 2i - 2sqrt{5} - 2}}{-4i - 2sqrt{5}i} ]
Further simplification leads to:
[ frac{pi}{sqrt{5}} e^{sqrt{3} - 2sqrt{5} - 2i} ]
Challenges and Considerations
The complexity of the result might suggest a more straightforward answer, especially if the integral yields a real value. However, the appearance of a complex solution does not necessarily mean the calculation is incorrect. In cases where the integral is originally real and requires a complex contour, the complex result is expected.
It's crucial to meticulously verify each step, particularly in complex analysis, as algebraic or conceptual mistakes can lead to incorrect results. For instance, evaluating a real-valued integral using a contour integral might not yield a straightforward real value, indicating a potential error in the initial setup or algebraic manipulation.
The result, though complex, remains valid within the framework of complex analysis. This exercise highlights the importance of rigorous problem-solving and validating each step carefully.