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Given $$\int_{\gamma}\frac{1}{(z-a)(z-\frac{1}{a})}dz,$$ and $0<a<1$, where $\gamma(t)=e^{it}$ and $0\le t \le 2\pi$

I am trying to find the residue of$$f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}$$

The answer says $\operatorname{res}(f,\mathbb C)=\frac{1}{(a-\frac{1}{a})}$?

Any help will be appreciated, thanks.

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Residues are defined at a pole of finite order. So, this is a function from the poles of a meromorphic function to the complex numbers. Are you attempting to compute an integral over a path from the residue – ncmathsadist Jan 19 '12 at 1:31
Is your ultimate goal to compute that integral? If so, why are you computing the residues at both singularities? Do both singularities contribute to the value of the integral? – Santiago Canez Jan 19 '12 at 1:46
@Santiago, Hi, yes my ultimate goal is to compute the integral. Sorry I just started learning residues so Im not good at these. So are you saying I can just take $\frac{1}{(a-\frac{1}{a})}$? How would I know which singularities contribute to the value of the integral? – Thomas Jan 19 '12 at 1:51
@Thomas, no need to apologize. Which version of the "Residue Theorem" are you trying to use? The answer to your problem is provided by ncmathsadist below, but if you're just starting out, you may not know what "winding number" means. – Santiago Canez Jan 19 '12 at 1:57
Just one more thing to think about: why is the restriction that 0 < a < 1 important? – Santiago Canez Jan 19 '12 at 1:59

There's also a simpler way of doing this. Observe that the function is holomorphic except at $z=a$ and $z=\dfrac{1}{a}$, then note your contour is the unit circle, as some other answers have discussed (if you're lost on this point, observe as a quick and easy method that generally contours with angles are circles or parts of circles, that we have $0$ to $2 \pi$ here, so clearly talking about one circle and you have $e^{it}$ which has modulus one so a unit radius). Now as $0 < a < 1$, $\frac{1}{a}$ is outside the circle so only need to look at $a$. You could use a Laurent series as previous answer has it, but this is quite messy (often one can avoid using Laurent's theorem). Now, there is a theorem (consult your textbook for a proof) that says a point is a pole of order m if $f(z)$ can be written as $\dfrac{\phi(z)}{{(z-z_0)}^m}$ where $\phi(z)$ is analytic and nonzero at $z_0$. Moreoever, the residue at $z_0$ is $\phi(z_0)$ if $m=1$ and $\dfrac{\phi^{m-1}(z_0)}{(m-1)!}$ if $m \geq 2$. So, here we note for $z=a$ we can set $\phi(z)=\dfrac{1}{z-1/a}$ and then we have that $\phi(a)=\dfrac{1}{a-1/a}$ and we note that $m=1$ in this case, so this is the residue and we are done.

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Are you familiar with the rule of residue theory that says that if $f(z)$ is analytic and has a simple zero at $z_0$, then Res$[\frac{1}{f(z)}, z_0]= \frac{1}{f'(z_0)}$? Try using this to evaluate the residue. I hope you get the same answer as mentioned in your post.

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Thanks, I just edited my original post, does it make sense now? – Thomas Jan 19 '12 at 1:37
One minor note here: In that rule I mentioned from residue theory, you would have to let the denominator of your original function become the new "$f(z)$." – Libertron Jan 19 '12 at 1:51
@Santiago: Although the OP doesn't explicitly mention computing an integral, that's probably what he must be trying to do after computing the residue I guess. – Libertron Jan 19 '12 at 1:53

Now consider the winding number of the curve about the two poles. This will do the job.

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$$ f(z)=\frac{1}{(z-a)(z-\frac{1}{a})}=\frac{\frac{1}{a-\frac{1}{a}}}{z-a}+\frac{\frac{-1}{a-\frac{1}{a}}}{z-\frac{1}{a}} $$ $$ \gamma \text{ is unit circle} \hspace{20mm} \text{Poles of Function} \begin{cases} z_1=a &\text{placed inside of unit circle} \\ z_2=\frac{1}{a} &\text{placed outside of unit circle} \end{cases} $$ $$ |a|<1 \rightarrow |\frac{1}{a}|>1 \hspace{10mm} $$ $$ \text{Newton's generalised binomial theorem} $$ $$ (a+b)^n =a^n+\frac{n}{1!}a^{n-1}b+\frac{n(n-1)}{2!}a^{n-2}b^2+...+b^n $$ $$ \frac{\frac{1}{a-\frac{1}{a}}}{z-a}= \frac{1}{a-\frac{1}{a}}(z-a)^{-1}=\frac{1}{a-\frac{1}{a}}(z^{-1}+z^{-2}a+z^{-3}a^2+...)=\frac{1}{a-\frac{1}{a}}\sum_{n-1}^{\infty}z^{-n}a^{n-1}$$ $$ \text{so residue is coefficient of term }z^{-1} \rightarrow\frac{1}{a-\frac{1}{a}} $$ $$ I=\int_{\gamma}\frac{1}{(z-a)(z-\frac{1}{a})}dz=2 \pi i .\operatorname{res}(f,\mathbb C)=\frac{2 \pi i}{(a-\frac{1}{a})} $$

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