Why the contour integral of $\,\displaystyle{f(z)=\frac{i-1}{z+i}}\,$ is not zero although it should be because $f(z)$ is analytic? I have used contour $z=\gamma(t)=2e^{it}$, where $0\leq t\leq\pi$.
Why the contour integral of $f(z)=\frac{i-1}{z+i}$ is not zero although it should be because $f(z)$ is analytic?
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complex-analysis
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1The function $f$ has a pole at $-i$. This is contained inside your circle of radius $2$. – 2012-08-25
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1Also: the specified contour is not closed. – 2012-08-25
1 Answers
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It makes no sense to say a function is analytic on its own - you must also specify an open subset of $\mathbb{C}$ that $f$ is analytic over. Here, $f$ is not analytic inside the specified contour - it has a simple pole at $z=-i$ which is contained inside.
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0But is satisfies C-R equations and first order partial derivatives($\frac{du}{dx},\frac{du}{dy},\frac{dv}{dx}$and $\frac{dv}{dy}$) are continuous. So it should be analytic. This converse of saying that if f is analytic then it satisfies C-R equations is not true. – 2012-08-25
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1It is analytic on the open set $\mathbb{C} \setminus \{-i\}$. What are conditions on the domain of analyticity which ensure that the contour integral of an analytic function is always zero ? – 2012-08-25
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1@alvoutila, it should be plain obvious to you that your function cannot be analytic within the circle $\,|z|\leq 2\,$ since $\,z=-i\,$ is an *obvious* pole of it there...isn't it? – 2012-08-25
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0So how you integrate so that integral goes to zero? which parametrization? – 2012-08-25
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1@alvoutila It actually doesn't satisfy the Cauchy-Riemann equations everywhere inside the contour. If you wish, we can write $z=x+iy$ so $f(x,y) = \frac{y-x-1}{x^2+ (y-1)^2} + i \frac{y+x-1}{x^2+(y-1)^2}.$ The function clearly isn't defined for $(x,y)=(0,1)$ so it's partials don't even exist there. The function is analytic everywhere but $z=i$ so if you choose any closed contour that does not enclose $i$ then your integral will be $0.$ – 2012-08-25
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1This integral of this function, around this contour, is not zero. It can be evaluated in terms of "residues", which you will presumably learn about soon in your course... – 2012-08-25
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0@Ahriman: domain must be simple closed contour with no singularities in it? – 2012-08-25
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1@alvoutila : Correct. I.e. with no singularities in the region that it surrounds. If it is holomorphic everywhere in that region and every where in some open set that the contour is in, then the integral is $0$. But this curve fails to be holomorphic at one point in the region that the curve surrounds. – 2012-08-25