# Prerequisites to Cauchy Integral Formula

1. Apr 1, 2017

### fishturtle1

1. The problem statement, all variables and given/known data
Calculate the integrals of the following functions on the given paths. Why does the choice of path change/not change each of the results?

(c) f(z) = exp(z) / z(z − 3)
https://www.physicsforums.com/file:///page1image10808 [Broken]
i. a circle of radius 4 centred at 0.

ii. a circle of radius 1 centred at 1 + i.
2. Relevant equations
(Cauchy’s Integral Formula). Suppose f is holomorphic in the region G and γ is a positively oriented, simple, closed, piecewise smooth path, such that w is inside γ and γ ∼G 0, then

2πif(w) = ∫f(z) / z-w dz

3. The attempt at a solution
I am confused whether or not I understand when to use Cauchy Integral Formula and when not to. I posted the definition that i'm working from in relevant equations.

i)

f(z) = ez / z(z-3)

1) I make my new f(z) = ez/z.
I say that my new f(z) is holomorphic in G, where G = ℂ\{0} because ez is entire and 1/z is holomorphic except at 0.

2) Then I checked if G enclosed 4eit (a circle centered at 0 with radius 4),
It does.

3) Then I check if 4eit contains w, where w = 3.
It does.

So after these 3 steps I believe i've satisfied all the conditions for the formula, and then I just plug in..

I = 2πif(w) = 2πi(e3/3) = (2πiee)/3

For the second one,

The path being integrated on is a circle of radius 1 centered at 1 + i.

I can't make w = 3 again, because then w won't be contained in the circle.

So I make f(z) = ez/ (z-3) and w = 0.

1) G = ℂ\{3}

2) G does enclose a circle of radius 1 centered at 1 + i.

3) w = 0, w is enclosed in a circle of radius 1 centered at 1 + i.

Then I evaluate, I = 2πif(w) = 2πi(1/-3) = -2πi / 3

I really just want clarification that I'm satisfying the requirements to using the formula, because there were some harder examples in class where we had to cut the path into parts and before i try to do that i want to make sure I can do these simpler looking problems.

Last edited by a moderator: May 8, 2017
2. Apr 1, 2017

### FactChecker

Apply the Cauchy Residue Theorem and figure out the winding numbers of the paths around singularities of the function.

3. Apr 2, 2017

### fishturtle1

I'm not trying to be rude but for the textbook I'm using, I am 4 chapters behind the Cauchy Residue Theorem. Is there another way to approach this problem without that theorem?

I did try to fix my first problem. I realized that it has 2 points of discontinuity so I split the path of integration.

I split the path of the integral into two semicircles... I think it is called partial decomposition..

so here's my work:

$\int_{C[0,4]} \frac {\frac {e^z} {z}} {z-3} dz + \int_\gamma f(z)dz + \int_{-\gamma} f(z)dz + \int_{C[0,4]} \frac {\frac{e^z} {z-3}} {z} dz$

= $\int_{C[0,4]} \frac {\frac {e^z} {z}} {z-3} dz + \int_{C[0,4]} \frac {\frac{e^z} {z-3}} {z} dz$

= $2\pi i \left[ \frac {e^w} {w} \right] _{w=3} + 2\pi i \left[ \frac {e^w} {w-3} \right] _{w=0}$

= $2\pi i[\frac {e^3} {3} - \frac {1} {3}]$

...

I haven't figured out the second one but i'm pretty sure my initial solution is wrong because it has no points of singularity inside it but I will look more at youtube/textbook for now, I just wanted to post something in the mean time. Is this what you meant by looking at the singularities ?

4. Apr 2, 2017

### Ray Vickson

Expand 1/(z(z-3)) in partial fractions and then evaluate the two integrals separately.

Last edited by a moderator: May 8, 2017
5. Apr 2, 2017

### fishturtle1

I think I did that here:

but I'm not sure how to approach the second one because their are no singularities inside the path