# Applying Cauchy Residue Theorem

For $C := \{z(t) : t(1+i) : t \in [-1,1]\}$, $\int_{C} \frac{dz}{(z-1)}$. The singularities of $\frac{1}{(z-1)}$ is $z_0 = 1$. Note that this singularity (pole?) is contained within the contour. Hence, by Cauchy Residue Theorem, $\int_{C} \frac{dz}{(1-z)} = 2\pi i(Res(f,z_0)) = 2\pi i(1) = 2\pi i$

Is this correct? Going through with long way (i.e, using the definition of complex integral), I get a different answer.

Can anyone either confirm if my answer is correct or point out what I have done wrong?

• Your $C$ is not closed. – mrf Apr 23 '15 at 7:13
• Hm, so by apply CRT to this when C is not closed, what does that mean geometrically? Or is it just wrong. Looking at the theorem statement again, I see that it is a necessary condition that $C$ be closed, but why? – Telon Apr 23 '15 at 7:17
• For one thing, the only way for a pole to be "inside" a contour is if the contour is a closed curve. – Christopher A. Wong Apr 23 '15 at 7:48