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
\int_{\Gamma} e^{1/z} dz for \Gamma any closed path not passing through the origin.
We need Res(e^{1/z}, 0)
It was found in a previous example that 0 is an essential singularity of e^{1/z}. There is no simple general formula for the residue of a function at an essential singularity. However,
e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!}\frac{1}{z^n}
is the Laurent expansion of e^{1/z} about 0, and the coefficient of 1/z is 1. Thus, Res(e^{1/z},0) = 1 and
\int_{\Gamma} e^{1/z} dz = i2\pi
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So, my problem is \int_{\Gamma} e^{\frac{2}{z^2}} dz, and gamma is the square with sides parallel to the axes and of length 3, centered at -i
What I did was say that e^\frac{2}{z^2} = \sum_{n=0}^\infty \frac{1}{n!}\frac{2}{z^{2n}}
and again 0 is an essential singularity of the function. The coefficient of \frac{2}{z^2} is 2, and therefore Res(e^{\frac{2}{z^2}},0) = 2, and so
\int_{\Gamma} e^\frac{2}{z^2} dz = i4\pi
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
\int_{\Gamma} e^{1/z} dz for \Gamma any closed path not passing through the origin.
We need Res(e^{1/z}, 0)
It was found in a previous example that 0 is an essential singularity of e^{1/z}. There is no simple general formula for the residue of a function at an essential singularity. However,
e^{1/z} = \sum_{n=0}^\infty \frac{1}{n!}\frac{1}{z^n}
is the Laurent expansion of e^{1/z} about 0, and the coefficient of 1/z is 1. Thus, Res(e^{1/z},0) = 1 and
\int_{\Gamma} e^{1/z} dz = i2\pi
---------------------------------------------------
So, my problem is \int_{\Gamma} e^{\frac{2}{z^2}} dz, and gamma is the square with sides parallel to the axes and of length 3, centered at -i
What I did was say that e^\frac{2}{z^2} = \sum_{n=0}^\infty \frac{1}{n!}\frac{2}{z^{2n}}
and again 0 is an essential singularity of the function. The coefficient of \frac{2}{z^2} is 2, and therefore Res(e^{\frac{2}{z^2}},0) = 2, and so
\int_{\Gamma} e^\frac{2}{z^2} dz = i4\pi
Is this correct?