# Thread: evaluating complex integrals (probably cauchy formula)

1. ## evaluating complex integrals (probably cauchy formula)

1.Evaluate the following integrals
$\displaystyle {\int _{\partial B_2(i)}}\frac{e^{i\zeta}}{\zeta ^{4} - 4\zeta ^2} d\zeta$

$\displaystyle {\int _{\partial B_3(-1)}}\frac{\zeta^{3}}{\zeta ^{2} + \zeta^2 -2} d\zeta$

$\displaystyle {\int _{\partial B_1(1)}}\frac{sin(\pi\zeta)}{\zeta ^{2} - 1} d\zeta$

6. let $\displaystyle z_0$ $\displaystyle \epsilon$ $\displaystyle B_r(z_0)$, let $\displaystyle r>o$ and let $\displaystyle f:B_r[z_0] \to \mathbb{C}$ be continous such that $\displaystyle f|_{B_r(z_0)}$ is holomorphic. Show that
$\displaystyle f(z) = \frac{1}{2\pi i} \int _{\partial B_r(z_0)} \frac{f(\zeta)}{\zeta - z}d\zeta$

for all $\displaystyle z \epsilon B_r(z_0)$
(hint: for $\displaystyle \theta \epsilon (0,1)$, apply the cauchy integral formula to $\displaystyle B_{\frac{r}{\theta}}(z_0) \ni z \mapsto f(\theta(z-z_0)+z_0)$; then let $\displaystyle \theta \to 1$.)

wow this was a lesson in the use of these symbols and latex or w/e
anyways, thats whats left of my assignment due this time tomorrow
if someone could help me get on track I'd appreciate it
#6 looks like the original cauchy statement but i think it's saying that the border may not be holomorphic

2. I got #6 myself
so only need 1, a/b/c now

3. Maybe some are unfamiliar with your notation like I was. Ok, let's assume it's what I think it is: $\displaystyle \partial B_2(i)$ means the boundary of a circular disc with center at $\displaystyle i$ and radius $\displaystyle 2$. Just never seen that notation before. If so then I'll write it as:

$\displaystyle \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz$

(using $\displaystyle \zeta$ makes it just unnecessarily cluttered)

So that contour encloses only the pole at the origin and you want to solve it using Cauchy's Integral Formula:

$\displaystyle f^{(n)}(w)=\frac{n!}{2\pi i}\mathop\oint\frac{f(z)}{(z-w)^{(n+1)}}dz$

where $\displaystyle f(z)$ is analytic in an on the contour. Then:

$\displaystyle \mathop\oint\limits_{D(i,2)}\frac{e^{iz}}{z^4-4z^4}dz=\oint\frac{\frac{e^{iz}}{(z^2-4)}}{(z-0)^2}dz=2\pi i\frac{d}{dz}\left(\frac{e^{iz}}{z^2-4}\right)\Bigg|_{z=0}=\pi/2$

Here's a check in Mathematica:

Code:
In[3]:=
NIntegrate[(Exp[I*z]/(z^4 - 4*z^2))*2*I*
Exp[I*t] /. z -> I + 2*Exp[I*t],
{t, 0, 2*Pi}]
N[2*Pi*I*D[Exp[I*z]/(z^2 - 4), z] /. z -> 0]

Out[3]=
1.5707963267949334 - 8.459899447643693*^-14*I

Out[4]=
1.5707963267948966
Do the other ones just like that but I bet you figured it out already.

4. ok, I kept getting confused at the denominator being 0 but I guess that only matters on the border?
and when I do part b) I get two different solutions depending on what I take as f
example:

$\displaystyle 2\pi i f(1) = \frac{2}{3}\pi i$ for $\displaystyle f(z) = \frac{z^3}{z+2}$
and
$\displaystyle 2\pi i f(-2) = \frac{16}{3}\pi i$ for $\displaystyle f(z) = \frac{z^3}{z-1}$

is there a reason why I wouldn't use one or the other? or can I not use this approach for some reason

5. Hey, the denominator is never zero. It's like asking if the denominator is zero for $\displaystyle \int_1^2 \frac{1}{x}dx$. Never gets ther right? Same dif with the integration path for the first one: $\displaystyle D(i,2)=\{z : z=i+2e^{it},\;0\leq t \leq 2\pi\}$. That never gets to the point $\displaystyle 0$ or $\displaystyle \pm 2$. For the second one:

$\displaystyle \mathop\oint\limits_{D(-1,3)}\frac{z^3}{2z^2-2}dz=\frac{1}{2}\left[\oint\frac{\frac{z^3}{z-1}}{z+1}dz+\oint\frac{\frac{z^3}{z+1}}{z-1}dz\right]$

Where the contours on the right side are now only going around each respective pole in accordance with the requirements of Cauchy's Integral Theorem (we can split up $\displaystyle D(-1,3)$ that way without problems right? Now:

$\displaystyle =\frac{1}{2}\left[2\pi i\left(\frac{z^3}{z-1}\Bigg|_{z=-1}+\frac{z^3}{z+1}\Bigg|_{z=1}\right)\right]$

$\displaystyle =\pi i(1)=\pi i$

Also . . . until you get good with these, check them numerically:

Code:
In[1]:=
NIntegrate[(z^3/(2*z^2 - 2))*3*I*Exp[I*t] /.
z -> -1 + 3*Exp[I*t], {t, 0, 2*Pi}]

Out[1]=
-2.220446049250313*^-16 + 3.1415926306916244*I
If the numerical results don't agree with the symbolic results, and you're pretty sure you've set up the numerical calculations correctly, look for an error in the symbolic results.

6. $\displaystyle \mathop\oint\limits_{D(-1,3)}\frac{z^3}{z^2 + z-2}dz$

I screwed up on the original the denominator only one of them is squared
I'll try a few things though to get this
and unfortunately I have no way of checking these numerically since in pure math we don't use mapl or matlab or mathematica or anything so I don't have those