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August 3rd, 2012, 02:09 AM   #1
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complex curve integral

let c be the unite circle |z|=1 oriented counterclockwise with center 0.find the value of integral $S=\oint_{C}^{ }\frac{e^{z}}{1-cosz}$=?
the answer is $4\pi i$
my solution: since z=0 is the residue of S with a pole of order 2 I tried to use the following idea that
[attachment=0:3vg277n4]id.png[/attachment:3vg277n4]
then
$\lim_{z\rightarrow \0 }(\frac{d}{dz}(\frac{z^{2}e^{z}}{1-cosz}))=\lim_{z\rightarrow \0 }(\frac{(2ze^z+z^2e^z)(1-cosz)+(sinz)(z^2e^z)}{(1-cosz)^{2}})$=?
now I have no idea what to do here because I forgot many things about limit theorem.
I know the answer of $\lim_{z\rightarrow \0 }(\frac{d}{dz}(\frac{z^{2}e^{z}}{1-cosz}))$ should be 2 then multiply by $2\pi i$ = $4\pi i$
thanks
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August 3rd, 2012, 05:08 AM   #2
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Re: complex curve integral

Quote:
 Originally Posted by mhhojati I know the answer of $\lim_{z\rightarrow 0 }(\frac{d}{dz}(\frac{z^{2}e^{z}}{1-cosz}))$ should be 2 . . .[/latex]
I think that the limit diverges. But I could be wrong . . .

 August 3rd, 2012, 08:50 AM #3 Member   Joined: Jan 2012 Posts: 57 Thanks: 0 Re: complex curve integral thank you for reply ,I found that there is also another way to finding the value of S using McLauren Series $\frac{1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...}{1-(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)}$ the answer is the coefficient of $\frac{1}{z}$ which will be 2 but it's quit hard to divide $1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+..$ by $1-(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)$ and get the coefficient of $\frac{1}{z}$
 August 3rd, 2012, 01:01 PM #4 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: complex curve integral The Laurent series for $\frac{e^{z}}{1-\cos(z)}$ is $\frac{2}{z^{2}}+\frac{2}{z}+\frac{7}{6}+\frac{z}{2 }+O(z^{2})$ As you pointed out, the coefficient of the 1/z term is your residue. It is 2. $2\pi i(2)=4\pi i$ I'd show some steps for the long division, but it is difficult to show that here. Yes, this kind of long division can be tedious. Just be extra careful. Make sure you line up like powers and be careful with signs. But, for that matter, you only need 2 steps to get the 1/z term.
August 4th, 2012, 01:17 AM   #5
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Re: complex curve integral

Quote:
 Originally Posted by galactus The Laurent series for $\frac{e^{z}}{1-\cos(z)}$ is $\frac{2}{z^{2}}+\frac{2}{z}+\frac{7}{6}+\frac{z}{2 }+O(z^{2})$
[color=#000000]Galactus, how did you compute the Laurent expansion?[/color]

 August 4th, 2012, 03:14 AM #6 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: complex curve integral Hey Z. I long divided by using the series for e^z and 1-cos(z). Also, to make sure I did not make a mistake while dividing (which is easy to do), I checked it with Maple. I would show my work, but I have no idea how to type up long division format with LaTex. You can see the first term, $\frac{2}{z^{2}}$ by looking at the problem. $\frac{1}{(\frac{z^{2}}{2})}=\frac{2}{z^{2}}$. Then, continuing on.
August 4th, 2012, 03:25 AM   #7
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Re: complex curve integral

[color=#000000]
Quote:
 Originally Posted by mathbalarka I think that the limit diverges. But I could be wrong . . .
No, you don't compute such residues by taking limits. These residues are computed with Laurent expansions. But in this case you could use the limit since z=0 is a simple pole.

Quote:
 Originally Posted by galactus You can see the first term, $\frac{2}{z^{2}}$ by looking at the problem. $\frac{1}{(\frac{z^{2}}{2})}=\frac{2}{z^{2}}$. Then, continuing on.
I thought you had some special way that I don't know, I don't see how you managed to find it (except from using maple) but here is my try and if I am wrong please correct me.

$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+\mathcal{O}(x^6)\Rightarrow \frac{1}{1-\cos(x)}=\frac{2}{x^2}\left(1-\frac{x^2}{12}+\mathcal{O}(x^{4})\right)^{-1}=\frac{2}{x^2}\left(1+\frac{x^2}{12}+\mathcal{O} (x^4)\right)=\frac{2}{x^2}+\frac{1}{6}+\mathcal{O} (x^4)$

so knowing that $e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}$

$\frac{e^{x}}{1-\cos(x)}=\left(\frac{2}{x^2}+\frac{1}{6}+\mathcal{ O}(x^4)\right)\left(1+x+\mathcal{O}(x^2)\right)=\f rac{2}{x^2}+\frac{2}{x}+2\mathcal{O}(1)+\frac{1}{6 }+\frac{x}{6}+\frac{1}{6}\mathcal{O}(x^2)+\mathcal {O}(x^4)+\mathcal{O}(x^5)+\mathcal{O}(x^6)=\frac{2 }{x^2}+\underset{\underset{c_{-1}=2}{\downarrow}}{\frac{2}{x}}+2\mathcal{O}(1)$

and so the residue is 2.[/color]

 August 4th, 2012, 03:55 AM #8 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: complex curve integral just use regular ol' algebra division. I just used the first few terms. I tried lining them up best I could, but it's kind of a pain. Regardless, once we see the 1/z term, that is all we really need. Code:  2/z^2 + 2/z +7/6+z/2+........ -------------------------------- z^2/2 - z^4/24) 1+z+z^2/2+z^3/6+z^4/24+...... 1 -z^2/12 --------------------------------- z + 7z^2/12 + z^3/6 + z^4/24 z -z^3/12 ------------------------------------ 7z^2/12 + z^3/4 + z^4/24 7z^2/12 - 7z^4/144 ----------------------------------------- z^3/4 + 13z^4/144 But, all we have to do is divide the series for 1-cos(z) into the series for e^z. $\frac{1+z+\frac{z^{2}}{2}+\frac{z^{3}}{6}+\frac{z^ {4}}{24}}{\frac{z^{2}}{2}-\frac{z^{4}}{24}$ This is all I used to get the first 4 terms.
 August 4th, 2012, 05:32 AM #9 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: complex curve integral [color=#000000]Ok.[/color]
August 6th, 2012, 12:59 AM   #10
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Re: complex curve integral

Quote:
 Originally Posted by mhhojati $\lim_{z\rightarrow \0 }(\frac{d}{dz}(\frac{z^{2}e^{z}}{1-cosz}))=\lim_{z\rightarrow \0 }(\frac{(2ze^z+z^2e^z)(1-cosz)+(sinz)(z^2e^z)}{(1-cosz)^{2}})$=? now I have no idea what to do here because I forgot many things about limit theorem. I know the answer of $\lim_{z\rightarrow \0 }(\frac{d}{dz}(\frac{z^{2}e^{z}}{1-cosz}))$ should be 2 then multiply by $2\pi i$ = $4\pi i$ thanks
[color=#000000]Well if f has a pole $z_0$ of m-th order then $\fbox{\textrm{Res}\left(f,z_0\right)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{\textrm{d}^{m-1}}{\textrm{d}z^{m-1}}\left[(z-z_0)^{m}f(z)\right]}$,

since $f(z)=\frac{e^{z}}{1-\cos(z)}$ has a simple pole at 0, m=1, so using the above formula
$\textrm{Res}\left(\frac{e^z}{1-\cos(z)};z=0\right)=\frac{1}{0!}\lim_{z\to 0 }\frac{ze^{z}}{1-\cos(z)}=\lim_{z\to 0 }\frac{ze^{z}}{1-\cos(z)}=\lim_{z\to 0 }\frac{(z+1)e^{z}}{\sin(z)}=\lim_{z\to 0 }\frac{(z+2)e^{z}}{\cos(z)}=\frac{(0+2)e^{0}}{1}=2$.[/color]

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