My Math Forum Mistake with a simple residue problem!

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 March 15th, 2012, 03:03 AM #1 Newbie   Joined: Jan 2011 Posts: 20 Thanks: 0 Mistake with a simple residue problem! (First I will forget the $2{\pi}i$ that is required when calculating every residue as it is not important here) Calculate the residue at the simple pole $z=-\frac{i}{2}$ for the function $f(z)=\frac{1}{(2z+i)(z+2i)}$ If I use "The Residue at a Pole" lemma which states: Let $\alpha$ be a pole of order $m$ of $f:A \subseteq {\it complex} \rightarrow {\it complex}$ so that $f(z)=\frac{g(z)}{z-\alpha}$, for some g, analytic at $\alpha$with $g(a) \neq 0$, then $Res(\alpha,f(z))= g(\alpha)$ For my function, I have simple poles (which is why I post a simpler an non-general version of the lemma), but when I apply this Lemma to find residue at my pole, I have: $Res(\frac{-i}{2},f(z))=g(\frac{-i}{2})=f(\frac{-i}{2})(z+\frac{-i}{2})$ This gives me: $Res(\frac{-i}{2},f(z))= \frac{1}{\frac{-i}{2}+2i}$ But this is incorrect! The correct answer is $Res(\frac{-i}{2},f(z))= \frac{1}{2(\frac{-i}{2}+2i)}$ Where is this 1/2 coming from?? And similarly, I face another problem: Find the residue at the point $z=\frac{-1}{3}$ for the function $f(z)=\frac{z^3}{(3z+1)^2(z+3)^2}$ At -1/3 I have a pole of order 2, so I rewrite $f(z)=\frac{z^3}{(z+3)^2}$, differentiate (because I do not have a simple pole, it is of order 2 => differentiate once) and insert in $z=-\frac{1}{3}$, which gives me $Res(-\frac{1}{3}, f(z))=\frac{13}{256}$ But this is also incorrect, as the correct value is $Res(-\frac{1}{3},f(z))= \frac{13}{2304}$, which is $\frac{13}{256} * \frac{1}{9}= \frac{13}{256} * \frac{1}{3^2}$... Where are these 1/2 and 1/9 terms coming in??
 March 15th, 2012, 07:19 AM #2 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: Mistake with a simple residue problem! [color=#000000]Ok you confused me! You have $f(z)=\frac{1}{(2z+i)(z+2i)}$, $\color{red}\bullet$ $\textrm{Res}\left(f(z);z=-\frac{i}{2}\right)=\lim_{z\to-\frac{i}{2}}\left(z+\frac{i}{2}\right)\frac{1}{2\l eft(z+\frac{i}{2}\right)(z+2i)}=\frac{1}{2i-\frac{i}{2}}=\frac{2}{3i}$ for the other one $f(z)=\frac{z^3}{(3z+1)^{2}(z+3)^{2}}$ you have a second order pole at z=-3. Use the formula $\lim_{z\to z_{0}}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}\left(\left((z-z_{0}\right)^{n}f(z)\right)$, in our case n=2 and $\color{red}\bullet$ $\textrm{Res}\left(f(z);z=-3\right)=\lim_{z\to-3}\frac{1}{1!}\frac{d}{dz}\left((z+3)^{2}\frac{z^3 }{(3z+1)^{2}(z+3)^{2}}\right)=\lim_{z\to -3}\frac{d}{dz}\frac{z^3}{(3z+1)^2}=\lim_{z\to -3}\frac{3z^2 (z+1)}{(3z+1)^{3}}=\frac{27}{256}$ for the pole at $z=-\frac{1}{3}$ write f in the following form and follow the above steps $f(z)=\frac{z^3}{3^2 \left(z+\frac{1}{3}\right)^{2}(z+3)^2}=\frac{z^3}{ 9 \left(z+\frac{1}{3}\right)^{2}(z+3)^2}$ .[/color]

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