# Thread: Cauchy Theorem Integration Part 2

1. ## Cauchy Theorem Integration Part 2

My book seems to do something like this here. My only question is that if we integrate with respect to theta...will it still work?

The reason I chose to do it this way is precisely because it is so obviously clear that this is $2 cos \theta$. But otherwise, it doesn't make a lot of sense to me this way.

2. After more thinking, I tried what seemed a little more natural:

$\int_\gamma \! (z + \dfrac{1}{z}) \, dz = \int_\gamma \! z \, dz + \int_\gamma \! z^{-1} \, dz = \dfrac{z^2}{2}\bigg|_a^b + Log \, z \big|_a^b$

$= \dfrac{(6+2i)^2 - (-4+i)^2}{2} + ln|6+2i| - ln|-4+i| + i[arctan(1/3) - arctan(-1/4)]$

This gives me a completely different answer.

3. Originally Posted by davismj
After more thinking, I tried what seemed a little more natural:

$\int_\gamma \! (z + \dfrac{1}{z}) \, dz = \int_\gamma \! z \, dz + \int_\gamma \! z^{-1} \, dz = \dfrac{z^2}{2}\bigg|_a^b + Log \, z \big|_a^b$

$= \dfrac{(6+2i)^2 - (-4+i)^2}{2} + ln|6+2i| - ln|-4+i| + i[arctan(1/3) - arctan(-1/4)]$

This gives me a completely different answer.
The $\arctan$ part is not correct. It depends on the sign of the real and imaginary term. You need the principal argument of the complex number $-4+i$:

$\int_{\gamma} z+1/z dz=17/2+16i+Log[6+2i]-Log[-4+i]$

$=(17/2+16i)+\ln|6+2i|-\ln|-4+i|+i\left(Arg(6+2i)-Arg(-4+i)\right)$

Also, you can compute this parametrically with a straight-line between the points:

$z=x(t)+iy(t)=\{t,i(1/10(t+4)+1)\}$

then we'd have:

$\int_{-4}^6 (z(t)+1/z(t))(1+1/10 t) dt\approx 8.9+13.4 i$