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We are going to be examined on using the Laurent series expansion to find integrals along simple closed curves. But the notes and lectures barely covered it and we only have 2 examples given.

Can someone explain how to solve a question like this:


Determine the Laurent series expansion of

$$f(z)=\frac{1}{1+z^2}$$ about $z_0=i$, that is valid in the region $0<|z-i|<2$ and hence evaluate

$$\int_\gamma \frac{1}{1+z^2}dz$$

where $\gamma(t)=i+e^{it}, 0\leq t\leq2\pi$


EDIT:

So is $$f(z)=\frac{1}{1+z^2}=-\frac{1}{2i}.\frac{1}{z+i}+\frac{1}{2i}.\frac{1}{z-i}$$ $$=\frac{1}{2i}.\frac{1}{z-i}-\frac{1}{2i}.\left(1-\frac{z-i}{2i}+\frac{{(z-i)}^2}{(2i)^2}...\right)$$

The Laurent series expansion.

And since this function has a single pole $i$ inside the circle, the coefficient of $\frac{1}{z-i}$ is the residue? So Res=$\frac{1}{2i}$

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  • $\begingroup$ Hint: Note that $1+z^2=(1+iz)(1-iz)$. By other hand, partial fractions can be usefull. $\endgroup$ – DiegoMath Apr 17 at 20:31
  • $\begingroup$ @DiegoMath hints arent very useful as I have no idea what's going on in this question. I already know how to use Cauchy's Integral formula to find integral of the two partial fractions which add together to give the original fraction. But I want to use the Laurent series expansion. $\endgroup$ – Mohamad Moustafa Apr 17 at 20:36
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Do a partial fraction decomposition : $\frac1{z^2+1}=\frac1{(z+i)(z-i)}=-\frac1{2i}(\frac1{z+i}-\frac1{z-i})$.

Then use the geometric series to get the Laurent series: $\frac1{z+i}=\frac1{z-i+2i}=\frac1{2i}(\frac1{1+\frac {z-i}{2i}})=\frac1{2i}\sum_{n=0}^\infty(-\frac {z-i}{2i} )^n$. This converges for $\mid z-i\mid\lt 2$.

So: $\frac1{2i}(\frac1{z-i}+\sum_{n=0}^\infty(-\frac {z-i}{2i} )^n)$ is the Laurent series centered at $i$.

Now use the residue theorem (the residue is $\frac1{2i}$) to compute the integral.

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  • $\begingroup$ I edited the question, can you please have a quick look to see if I understood correctly? And thanks alot for that answer. $\endgroup$ – Mohamad Moustafa Apr 18 at 0:58
  • $\begingroup$ I think you have it. Except we lost a minus sign. $\endgroup$ – Chris Custer Apr 18 at 1:04
  • $\begingroup$ Why did you expand the 1/z+i and not 1/z-i instead?Is it because we already know the coefficient of (z-i)^-1 in 1/z-i but not in 1/z+i? $\endgroup$ – Mohamad Moustafa Apr 18 at 1:05
  • $\begingroup$ The latter is already in the right form. We want to expand about $i$. So we want powers of $(z-i)$. $\endgroup$ – Chris Custer Apr 18 at 1:07
  • $\begingroup$ Ok. Sorry but I have another question. For 1/[(z+1)(z+3)] if we want a Laurent expansion valid for 1<|z|<3. How would I need to expand that? I can do the work myself but what should I turn this into? It seems this has two poles. $\endgroup$ – Mohamad Moustafa Apr 18 at 1:12
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observe that $z^2+1=(z-i)^2+2i(z-i)$. You can then look for the laurent series of $\frac{1}{y^2+2iy}$ about $y=0$.

You can show that $\frac{1}{(2iy)(1+\frac{y}{2i})}=\frac{1}{2iy}(1+\frac{y}{2i})^{-1}=\frac{1}{2iy}(1-\frac{y}{2i}+O(y^2))=\frac{1}{2iy}+\frac{1}{4}+\frac{1}{8}iy+O(y^2).$

Now switch $y$ and $z-i$, you get: $\frac{-i}{2(z-i)}+\frac{1}{4}+\frac{i(z-i)}{8}+O(z^2).$

You can now find the residue, as it corresponds to the coefficients of $(z-i)^{-1}$. You might want to use Cauchy-integral formula as a sense check at the end.

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  • $\begingroup$ what's O(y^2)? I think we are supposed to use binomial and Taylor expansions to find the Laurent series expansion. $\endgroup$ – Mohamad Moustafa Apr 17 at 22:15
  • $\begingroup$ $O(y^2)$ just means all terms where the power of y is greater than or equal to 2. if it confuses you then you might want to replace it with ... (dot,dot,dot). I've used nothing more than the expansion $(1+x)^-1=1-x+x^2+....$ I guess that should be okay. $\endgroup$ – Alexandros Apr 17 at 22:39
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You have a function $f(z)$ that has two poles, $z=\pm i$, why? because you can write your function as: $$ f(z)=\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} $$

If we restrict ourselves to the domain consisting of the circle of radius $1$ centered at $i$ then your function can be writen as: $$ f(z)=\frac{g(z)}{z-i} $$

where $g(z)=\frac{1}{z+i}$ is an analitic function in that domain and so admits a taylor series of the form: $$ g(z)=\sum_{n=0}^\infty a_n(z-i)^n $$

and so the Laurent series of $f$ can be expresed as:

$$ f(z) = \sum_{n=0}^\infty a_n(z-i)^{n-1} $$ where the $a_n$ are the coefficients from the Taylor series of the function $g$.

This coefficients can be computed as any other Taylor series: $$a_0 = g(i) = \frac{1}{2i}$$ $$a_n = \frac{1}{n!}\frac{d^ng(i)}{dz^n}$$

that is, near $z=i$ the function $g$ can be expanded as

$$ g(z) = \frac{1}{2i} + \frac{1}{4}(z-i)+\dots $$

and so the function $f$

$$ f(z) = \frac{1}{2i(z-i)} + \frac{1}{4}+\dots $$

If we now want to compute the integral of $f$ around $\gamma$ we have that:

$$ \int_\gamma f(z)\mathrm{d}z =\int_\gamma \left(\frac{1}{2i(z-i)} + \frac{1}{4}+\dots\right)\mathrm{d}z = \int_\gamma \frac{1}{2i(z-i)}\mathrm{d}z + \int_\gamma\frac{1}{4}\mathrm{d}z+\dots =\frac{1}{2i}\int_\gamma \frac{1}{(z-i)}\mathrm{d}z $$

as all the other integrands are analitic in the whole circle so a closed intregal vanishes. This result could also be writen from the begining taking into account that $g$ is analitic as:

$$ \int_\gamma f(z)\mathrm{d}z =\int_\gamma \frac{g(z)}{z-i}\mathrm{d}z=g(i)\int_\gamma \frac{1}{z-i}\mathrm{d}z $$

Finally taking into account that $$ \int_\gamma \frac{1}{z-i}\mathrm{d}z = \int_{\mathrm{B}_1(0)} \frac{1}{z}\mathrm{d}z=2\pi i $$

We have that

$$ \int_\gamma f(z)\mathrm{d}z = 2\pi i g(i)= \frac{2\pi i}{2i} = \pi $$

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