Find the Laurent series expansion of the following function and hence find the integral over the unit circle centred at i

Question

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:

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

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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=$\frac{1}{2i}$

Answer 1

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.

Answer 2

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.

Answer 3

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

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