# Cauchy's integral formula help

• May 9th 2010, 09:28 AM
EoinCahill
Cauchy's integral formula help
Q: Using Cauchy’s integral formula, compute the integral of g(z) over the circle of radius 3 centred at the origin, the contour integral being taken counterclockwise.

g(z) =z^3 + z^2 − 5/z − 2

i cant wrap my head around this formula or how to even begin solving it. all other examples had two 'poles' rather tha just (z-2)

any help will be much appreciated.
• May 9th 2010, 09:50 AM
Bruno J.
Can you set up the integral?
What does Cauchy's integral formula tell you?

By the way, it's not clear if you meant $z^3+z^2-5/z -2$ or $z^3+z^2-\frac{5}{z -2}$.
• May 9th 2010, 10:08 AM
EoinCahill
sorry im not sure what you mean by set up.
to be clear its (z^3+z^2-5)/(z-2)
I know i have to use the denominator to find the singularities of g(z) but i do not know how to plug in the equation into the formula.
• May 9th 2010, 11:05 AM
Bruno J.
Quote:

Originally Posted by EoinCahill
sorry im not sure what you mean by set up.
to be clear its (z^3+z^2-5)/(z-2)
I know i have to use the denominator to find the singularities of g(z) but i do not know how to plug in the equation into the formula.

That's a long shot from what you had written. You do know that $a+b \times c \neq (a+b) \times c$, right? Parentheses are not a luxury.

It's clear that the only singularity is at $z=2$.
Now can you write $g(z)dz$ in the form $\frac{f(z)}{z-2}dz$, where $f$ is holomorphic inside the contour? What does Cauchy's formula tell you about the integral of such a form along the contour?
• May 10th 2010, 03:39 AM
EoinCahill
Quote:

It's clear that the only singularity is at z=2.
yes i thought that myself.
so f(z) = z^3+z^2-5??

that the integral is equal to 2.pi.i??
• May 10th 2010, 03:43 AM
mr fantastic
Quote:

Originally Posted by EoinCahill
yes i thought that myself.

so f(z) = z^3+z^2-5??

that the integral is equal to 2.pi.i??

Yes. No.

You need to review Cauchy's Integral formula (see post #4 here for example (in your question n = 1): http://www.mathhelpforum.com/math-he...uchys-thm.html)
• May 10th 2010, 07:28 AM
EoinCahill
ok this is what i have so far.

f(z)= ( z^3+z^2-5 dz
------) z-2

is this right???
• May 10th 2010, 05:31 PM
mr fantastic
Quote:

Originally Posted by EoinCahill
ok this is what i have so far.

f(z)= ( z^3+z^2-5 dz
------) z-2

is this right???

I said in my earlier reply that yes your f(z) was correct and no your answer was not correct. Where has the above come from - it is nothing more than a re-statement of the question! Did you read the post I refered you to? You need to compare the given integral to Cauchy's Integral Formula and:

1. Identify f(z).

2. Identify $\alpha$.

Then you need to:

3. Evaluate $f(\alpha)$.