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August 3rd, 2012, 02:09 AM  #1 
Member Joined: Jan 2012 Posts: 57 Thanks: 0  complex curve integral
let c be the unite circle z=1 oriented counterclockwise with center 0.find the value of integral =? the answer is 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 =? now I have no idea what to do here because I forgot many things about limit theorem. I know the answer of should be 2 then multiply by = thanks 
August 3rd, 2012, 05:08 AM  #2  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: complex curve integral Quote:
 
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 the answer is the coefficient of which will be 2 but it's quit hard to divide by and get the coefficient of 
August 3rd, 2012, 01:01 PM  #4 
Senior Member Joined: May 2011 Posts: 501 Thanks: 6  Re: complex curve integral
The Laurent series for is As you pointed out, the coefficient of the 1/z term is your residue. It is 2. 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  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: complex curve integral Quote:
 
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 1cos(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, by looking at the problem. . Then, continuing on. 
August 4th, 2012, 03:25 AM  #7  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: complex curve integral [color=#000000] Quote:
Quote:
so knowing that 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 1cos(z) into the series for e^z. 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: 1,990 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  
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: complex curve integral Quote:
since has a simple pole at 0, m=1, so using the above formula .[/color]  

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