Answer to Question #146560 in Complex Analysis for Usman

Question #146560
Q: Evaluate the following integral using residue theorem
∫ Z^2 e^z dz ; C : |z| = 1 ,Answer(πi/3)
1
Expert's answer
2020-11-25T17:04:14-0500

"\\int_C z^2 e^z dz" , "C: |z| = 1"


The residue theorem:

"\\int_C f(z)dz = 2\\pi i \\sum_{k=1}^m res_{z= a_k} f(z)", where "a_k" - singularity point.


So we have found in which poin function has a singularity:

"f(z) = z^2 e^z," has a singularity only in "z = \\infin"


In our case, for "z = \\infin" formula from residue theorem will take the next form:

"\\int_C f(z)dz = -2\\pi i \\sum_{k=1}^m res_{z= a_k} f(z)"


So we have to find "res_{z=\\infin} f(z)"

Also we know "res_{z=\\infin} f(z)= - C_{-1}", where "C_{-1}" - coefficient near "z^{-1}"


Find the Laurent series of "f(z)". Series in negative power of z:

"f(z)=z^2e^z = z^2(\\sum_{n=0}^\\infin \\frac{1}{n!z^n}) = z^2(1+\\frac{1}{z} + \\frac{1}{2!z^2} + ... +\\frac{1}{n!z^n}+...)=\\\\= z^2 + z + \\frac{1}{2} + \\frac{1}{3!z}+ ...+ \\frac{1}{n!z^{n-2}}+..."


We can see that "C_{-1} = \\frac{1}{3!} = \\frac{1}{6}", so "res_{z=\\infin} f(z) = -\\frac{1}{6}" and

"\\int_C z^2 e^z dz = -2\\pi i (-\\frac{1}{6}) = \\frac{\\pi i}{3}"


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