# Why the contour integral of $f(z)=\frac{i-1}{z+i}$ is not zero although it should be because $f(z)$ is analytic?

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 $0\leq t\leq\pi$.

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The function $f$ has a pole at $-i$. This is contained inside your circle of radius $2$. –  fretty Aug 25 '12 at 12:54
Also: the specified contour is not closed. –  Sean Eberhard Aug 25 '12 at 15:14
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.
It 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 ? –  Ahriman Aug 25 '12 at 13:04
@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? –  DonAntonio Aug 25 '12 at 13:05
@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 $0.$ –  Ragib Zaman Aug 25 '12 at 13:10
@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. –  Michael Hardy Aug 25 '12 at 13:40