Complex countour integrals

[squaremeplz]

Homework Statement

1) evaluate the contour integral

c.int [ e^z / ((z+1)^2 * (z-4)) ] dz

about the rectangle with verticies -3 + 4i, 3 + 4i, -3-4i, 3 - 4i,

use partial fractions and the cauchy integral formula to evaluate the integral.

2) Let C be the closed contour that runs counterclockwise around the circle | z - 2 + i | = 3

compute c.int [z - 3z^2]dz

(the z raised to the first power in this equation is barred. i.e must integrate the complex conjugate of the first z.)

Homework Equations

cauchy integral formula f(z_0) = 1/(2*pi*i) * int [ f(z) / ( z - z_o) ] dz

The Attempt at a Solution

1. First I expanded e^z / ((z+1)^2 * (z-4)) using partial fraction decomposiion and got

e^z * ( (-1/25)/(z+1) + (-1/5)/(z+1)^2 + (1/25)/(z-4) )

brake up the integral into 3 parts

-1/25 * c.int [ e^z / (z + 1) ] dz - 1/5 * c.int [ e^z / (z + 1)^2 ]

+ 1/25 * c.int [ e^z / (z - 4) ]

applying the cauchy integral formula I get:

-1/25 *2 *pi * i * e^(-1) - 1/5 *2 *pi * i * e^(-1) + 0

the last term becomes zero because 4 lies outside the rectangle.

The only part about this that makes me uncomfortable is not using the vertecies of the rectangle at all in the computation of the curve. Did I need to parameterize the rectangle? Thanks

2) for the integral c.int [z - 3z^2]dz i parametrized the circle | z - 2 + i | = 3 as

z(t) = 2 + i + 3e^(it)
z'(t) = i3e^(it)

then evaluated c.int [ f(z(t))z'(t)dt ]

could someone please verify my methods.

Thanks!

 

[Billy Bob]

Mostly correct, good job!

Notice that for problem 1, the integral c.int [ e^z / (z + 1)^2 ] requires the CIF for derivatives, not the CIF you stated, but I believe the result is the same (double check it). Definitely don't parametrize the rectangle. The answer would be the same for any simple closed contour that encloses -1 but not 4, and that's part of what this question is testing that you know.

For problem 2, the center of the circle is 2-i, not 2+i (unless the typo is in the question instead of your solution). Also, you could make it somewhat easier by breaking the integral into (integral of z bar dz) - (integral of 3z^2 dz), where the second integral is very easy, thus simplifying the amount of integration that has to be calculated "by hand."

 

[latentcorpse]

quickly done using the residue theorem i get $-\frac{12 \pi i}{25 e}$