Evaluate the Integral

[Guest]

Here is one more two part question that is really frustrating me. Can someone walm me through them?

Evaluate the integral of the given function along the respective path. In the case of a simple closed path assume the positive orientation.
a. $\displaystyle f(z)=\frac{3}{z}-\frac{2}{z-2i},|z-2i|=1.$
b. $\displaystyle f(z)=\frac{z}{z^2-1},|z-3|=1.$



Evaluate the given definite integral.
a. $\displaystyle \int{_0^\pi}z cos(z^2)dz$

b. $\displaystyle \int{_0^\pi}sin^2zdz$


Thank You in advance.

 

[pka]

The second set of integrals is trivial. Both are analytic, thus their integrals are path-wise independent. So treat them as you did in first year calculus.

As for the first set, note that both contours are circles
In each case, singularities of the integrands may or may not be interior to the simple closed curve. Therefore, you must apply the Cauchy-Goursat Theorem.
You need to do that for yourself in order to learn it.

 

[Guest]

Ok...I see what you are saying here and I am looking at the Cauchy-Goursat Theorem and things are still not making sense. I have about a "million" of these to do, so I can be walked through the first two I should be good


$\displaystyle f(z)=\frac{3}{z}-\frac{2}{z-2i},|z-2i|=1$


$\displaystyle f(z)=\frac{z}{z^2-1},|z-3|=1$

 

[pka]

The contour $\displaystyle |z - 3| = 1$ is a circle with radius 1 and center (3,0). The singularities of $\displaystyle \frac{z}{{z^2 - 1}}$ are (1,0) & (-1,0) both exterior to that circle. What is the integral?


The contour $\displaystyle |z - 2i| = 1$ is a circle with radius 1 and center (0,2). Where are the singularities?

 

[Guest]

the integral

Would the integral be $\displaystyle \frac{1}{3}x(-3+x^2)$ right?...but the singularites portion...? Where are we going with this?