Complex Integral over a rectangular contour

Let C be a rectangle with corners $\pm 2\pm 3i$ in the anticlockwise direction. Find $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$

I tried:

Split the contour in to 4 parts: $C_1,C_2,C_3,C_4$.

Let's say $C_1$ is the the bottom side of the rectangle. So,

$C_1: f(t) =t-3i, -2\le t\le 2$

Then I find $$\int_{-2}^{2}\frac{1}{t-3i}+\frac{1}{t-3i-1}dt$$

Similarly, I would find the integrals over $C_2, C_3$ and $C_4$. Then, the final answer is the sum of $C_1,C_2,C_3,C_4$.

Is this correct?

• It's much simpler to use the Residue Theorem. – Robert Israel Feb 25 '17 at 1:27
• Your first integral has a lower limit of $c$ and no upper limit – mrnovice Feb 25 '17 at 1:27
• @mrnovice that is standard notation... – Simply Beautiful Art Feb 25 '17 at 1:29
• Oh ok, this is a topic beyond me then – mrnovice Feb 25 '17 at 1:29
• The two poles are inside the contour, just use the residue theorem. – Zaid Alyafeai Feb 25 '17 at 1:29

Let $C$ be a closed rectifiable contour with winding number $1$ about a point $z_0\in \mathbb{C}$. Cauchy's Integral Formula states that if $f$ is analytic on the open region enclosed by $C$ and continuous on $C, then $$f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}\,dz \tag 1$$ In$(1)$, set$f(z)=1$,$z_0=0$and$C$to be the rectangle defined in the OP. Then from$(1)$we have $$1=\frac{1}{2\pi i}\oint_C \frac{1}{z}\,dz$$ whereupon solving for$\oint_C \frac{1}{z}\,dz$reveals $$\oint_C \frac{1}{z}\,dz=2\pi i \tag 2$$ Similarly, set$f(z)=1$and$z_0=1$. Then from$(1)$we have $$1=\frac{1}{2\pi i}\oint_C \frac{1}{z-1}\,dz$$ whereupon solving for$\oint_C \frac{1}{z}\,dz$reveals $$\oint_C \frac{1}{z-1}\,dz=2\pi i \tag3$$ Putting together$(2)$and$(3)$yields $$\oint_C \left(\frac1z+\frac1{z-1}\right)\,dz=4\pi i$$ • The integral I'm looking for is $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$, so shouldn't the correct answer be$4\pi i$? Also, do the corners of the rectangle not matter at all when using the cauchy integral formula? – sucksatmath Feb 25 '17 at 14:44 • Yes, you're correct. It's$4\pi i\$. The corners don't matter. The contour need only be rectilinear. – Mark Viola Feb 25 '17 at 15:45