How to Use Cauchy Integral Formula for Homework Problems?

[jjangub]

Homework Statement

1) Evaluate [tex]\int[/tex]c ((5z-2)/(z(z-1)(z-3)))dz where c is the circle of radius 2 about the origin.
2) Evaluate [tex]\int[/tex]c (2*(z^2)-z+1) / ((z-1)^2(z+1))dz where c proceeds around the boundary of the figure eight formed by two circles of radius 1 with centres 1 and -1 by starting at 0, going once counterclockwise around the right circle followed by going once counterclockwise around the left circle.

Homework Equations

The Attempt at a Solution

1) I used partial fractions and got (-2)/(3z) + ((-3)/2(z-1)) + (13/6(z-3))
it has three isolated sigularities z = 0, z = 1, z = 3, only two are interior to c.
Since (-2)/(3z) is already a Larurent Series when 0 < lzl < 1 and
((-3)/2(z-1)) is Laurent Series when 0 < lz - 1l< 1.
therefore, (2*pi*i) * (-2/3) + (2*pi*i) * (-3/2) = (-13*pi*i) / 3

2) I used partial fractions and got 1/(z+1) + 1/(z-1) + 1/((z-1)^2). We have to use Caucy Integral Formula(CIF).
for 1/(z+1), multiply top and bot by (z-1) then (z-1)/((z+1)(z-1)). To use CIF, f(z) = (z-1)/(z+1) and f(z0) = 1, therefore (2*pi*i) * f(1) = 0
for 1/(z-1), multiply top and bot by (z+1) then (z+1)/((z+1)(z-1). To use CIF, f(z) = (z+1)/(z+1) and f(z0) = 1, therefore (2*pi*i) * f(1) = 2*pi*i
for 1/((z-1)^2), I don't know about this one, if I do this like other two terms, then I get
something/0.

This probably won't make sense at all, but I tried...
Please tell me if I did something wrong.
Thank you.

 

[vela]

Homework Statement

1) Evaluate [tex]\int[/tex]c ((5z-2)/(z(z-1)(z-3)))dz where c is the circle of radius 2 about the origin.
2) Evaluate [tex]\int[/tex]c (2*(z^2)-z+1) / ((z-1)^2(z+1))dz where c proceeds around the boundary of the figure eight formed by two circles of radius 1 with centres 1 and -1 by starting at 0, going once counterclockwise around the right circle followed by going once counterclockwise around the left circle.

Homework Equations

The Attempt at a Solution

1) I used partial fractions and got (-2)/(3z) + ((-3)/2(z-1)) + (13/6(z-3))
it has three isolated sigularities z = 0, z = 1, z = 3, only two are interior to c.
Since (-2)/(3z) is already a Larurent Series when 0 < lzl < 1 and
((-3)/2(z-1)) is Laurent Series when 0 < lz - 1l< 1.
therefore, (2*pi*i) * (-2/3) + (2*pi*i) * (-3/2) = (-13*pi*i) / 3

This is correct.

2) I used partial fractions and got 1/(z+1) + 1/(z-1) + 1/((z-1)^2). We have to use Caucy Integral Formula(CIF).
for 1/(z+1), multiply top and bot by (z-1) then (z-1)/((z+1)(z-1)). To use CIF, f(z) = (z-1)/(z+1) and f(z0) = 1, therefore (2*pi*i) * f(1) = 0
for 1/(z-1), multiply top and bot by (z+1) then (z+1)/((z+1)(z-1). To use CIF, f(z) = (z+1)/(z+1) and f(z0) = 1, therefore (2*pi*i) * f(1) = 2*pi*i
for 1/((z-1)^2), I don't know about this one, if I do this like other two terms, then I get
something/0.

This probably won't make sense at all, but I tried...
Please tell me if I did something wrong.
Thank you.

You're making this one too complicated. Look at it like this:

[tex]\int_C \frac{2z^2-z+1}{(z-1)^2(z+1)}\,dz = \int_C \frac{1}{z+1}\,dz + \int_C \left[\frac{1}{z-1}+\frac{1}{(z-1)^2}\right]dz[/tex]

The first integral on the RHS is proportional to the residue of the integrand at z=-1; similarly, the second integral on the RHS is proportional to the residue of the integrand at z=1. They're both conveniently in the form of Laurent series, so you can read off what b₁ is.

 

[jjangub]

so do I get zero as the answer?
I tried residue theorem and I got zero.

 

[vela]

I got 4πi. Show your work.

 

[jjangub]

For the first term, it has a taylor series of 1-z+z^2-z^3+...
and the coefficient of z is the desired residue, 2[tex]\pi[/tex]*i*(-1) = -2[tex]\pi[/tex]*i
For the second term, taylor series is z+2*z^2+3*z^3+...
and the coefficient of z is the desired residue, 2[tex]\pi[/tex]*i*(1) = 2[tex]\pi[/tex]*i
so if I add these two then, I get zero...
( I don't know why pi appears like that, but its just 2*pi*i)

 

[vela]

You're calculating the residue incorrectly. You have to expand the functions about the pole z₀, and the residue will be the coefficient of the 1/(z-z₀) term.

 

[jjangub]

I just learned residue today...is there any other way to do this?
well, I can use residue, but I am not confident about it.
I have to do this by tomorrow...

 

[vela]

It's simple. If you have a pole at z=z₀ inside the contour, expand the integrand in powers of (z-z₀). The coefficient of (z-z₀)^-1 is the residue.

For example, in the integral with the figure-8 contour

[tex]\int_C \frac{1}{z+1}\,dz[/tex]

there is a pole at z=-1, so you want to expand in powers of (z-(-1)) = z+1. Conveniently enough, the integrand is already in that form. (It's a Laurent series with only one term.) The coefficient of (z+1)^-1 is 1, so the integral is equal to

[tex]\int_C \frac{1}{z+1}\,dz = 2\pi i (1) = 2\pi i[/tex]

(If the contour encircled the pole going clockwise, you'd flip the sign.)