- #1
ColdFusion85
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First question pertains to the Residue Theorem
We are to use this theorem to evaluate the integral over the given path...
There is one problem from this section that I am stuck on. An example in the book evaluates
[tex]\int_{\Gamma} e^{1/z} dz [/tex] for [tex]\Gamma[/tex] any closed path not passing through the origin.
We need [tex]Res(e^{1/z}, 0)[/tex]
It was found in a previous example that 0 is an essential singularity of [tex]e^{1/z}[/tex]. There is no simple general formula for the residue of a function at an essential singularity. However,
[tex]e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!}\frac{1}{z^n}[/tex]
is the Laurent expansion of [tex]e^{1/z}[/tex] about 0, and the coefficient of 1/z is 1. Thus, Res([tex]e^{1/z}[/tex],0) = 1 and
[tex]\int_{\Gamma} e^{1/z} dz = i2\pi[/tex]
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So, my problem is [tex]\int_{\Gamma} e^{\frac{2}{z^2}} dz [/tex], and gamma is the square with sides parallel to the axes and of length 3, centered at -i
What I did was say that [tex]e^\frac{2}{z^2} = \sum_{n=0}^\infty \frac{1}{n!}\frac{2}{z^{2n}}[/tex]
and again 0 is an essential singularity of the function. The coefficient of [tex]\frac{2}{z^2}[/tex] is 2, and therefore Res([tex]e^{\frac{2}{z^2}}[/tex],0) = 2, and so
[tex]\int_{\Gamma} e^\frac{2}{z^2} dz = i4\pi[/tex]
Is this correct?
We are to use this theorem to evaluate the integral over the given path...
There is one problem from this section that I am stuck on. An example in the book evaluates
[tex]\int_{\Gamma} e^{1/z} dz [/tex] for [tex]\Gamma[/tex] any closed path not passing through the origin.
We need [tex]Res(e^{1/z}, 0)[/tex]
It was found in a previous example that 0 is an essential singularity of [tex]e^{1/z}[/tex]. There is no simple general formula for the residue of a function at an essential singularity. However,
[tex]e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!}\frac{1}{z^n}[/tex]
is the Laurent expansion of [tex]e^{1/z}[/tex] about 0, and the coefficient of 1/z is 1. Thus, Res([tex]e^{1/z}[/tex],0) = 1 and
[tex]\int_{\Gamma} e^{1/z} dz = i2\pi[/tex]
---------------------------------------------------
So, my problem is [tex]\int_{\Gamma} e^{\frac{2}{z^2}} dz [/tex], and gamma is the square with sides parallel to the axes and of length 3, centered at -i
What I did was say that [tex]e^\frac{2}{z^2} = \sum_{n=0}^\infty \frac{1}{n!}\frac{2}{z^{2n}}[/tex]
and again 0 is an essential singularity of the function. The coefficient of [tex]\frac{2}{z^2}[/tex] is 2, and therefore Res([tex]e^{\frac{2}{z^2}}[/tex],0) = 2, and so
[tex]\int_{\Gamma} e^\frac{2}{z^2} dz = i4\pi[/tex]
Is this correct?