Q1P - Mathematical Methods in Physical Sciences

Question

Let f(z)  be expanded in the Laurent series that is valid for all  z outside some circle, that is, $\left|z\right|\mathbf{>}\mathit{M}$(see Section 4). This series is called the Laurent series "about infinity." Show that the result of integrating the Laurent series term by term around a very large circle (of  radius > M) in the positive direction, is  $\mathbf{2}\mathit{\pi }\mathit{i}{\mathit{b}}_{\mathbf{1}}$ (just as in the original proof of the residue theorem in Section 5). Remember that the integral "around $\mathbf{\infty }$   " is taken in the negative direction, and is equal to  $\mathbf{2}\mathit{\pi }\mathit{i}$ : (residue at  $\mathbf{\infty }$ ). Conclude that $\mathit{R}\mathbf{(}\mathbf{\infty }\mathbf{)}\mathbf{=}\mathbf{-}{\mathit{b}}_{\mathbf{1}}$ . Caution: In using this method of computing $\mathit{R}\mathbf{(}\mathbf{\infty }\mathbf{)}$  be sure you have the Laurent series that converges for all sufficiently large  z.

Step-by-Step Solution

Answer

It is proved that the result of integrating Laurent series about infinity in the positive (contour clockwise) direction is $-2\pi i\left(b_1\right)$.

Step 1: Concept of Laurent series and Taylor series:

Laurent series expansion of  f(z)  about  z  = 0 is as follows:

 $f\left(z\right)=a_0+a_{1}z+a_2{z}^2+...+\frac{b_1}{z}+\frac{b_2}{z^2}+\frac{b_3}{z^3}+...$

Taylor series consist of   terms then:

 $f\left(z\right)=a_0+a_{1}z+a_2{z}^2+...+a_r{z}^r+\frac{b_1}{z}+\frac{b_2}{z^2}+\frac{b_3}{z^3}+...$

Residue theorem:

 ${\oint }_{c}f\left(z\right)dz=2\pi i\left(Ref\left(\infty \right)\right)$

Step 2: Simplify the function

Now Laurent series expansion of  f(z) about  z = 0 is,

 $f\left(z\right)=a_0+a_{1}z+a_2{z}^2+...+\frac{b_1}{z}+\frac{b_2}{z^2}+\frac{b_3}{z^3}+...$

 

So, obtain values as follows:

 $$\begin{gathered} a_n=\frac{1}{2\pi i}{\int }_c\frac{f\left(z\right)}{{\left(z\right)}^{n+1}}dz \\ {b}_n=\frac{1}{2\pi i}{\int }_c\frac{f\left(z\right)}{{\left(z\right)}^{-n+1}}dz \end{gathered}$$

Here,  C is a closed curve in the region  R and  $M<\left|z\right|<N$.

The Laurent series expansion of  f(z) consist of series of positive powers of  z, or Taylor series, and inverse power series of  z.

 

The Taylor series converge for $\left|z\right|<N$  and inverse power series converges for $\left|z\right|>M$ .

 

Let the outer radius  N is infinity. Then the series  f(z)  is convergent if the Taylor series consist of finite number of terms.

 

Let the Taylor series consist of   terms:

 $f\left(z\right)=a_0+a_{1}z+a_2{z}^2+...+a_r{z}^r+\frac{b_1}{z}+\frac{b_2}{z^2}+\frac{b_3}{z^3}+...$

Step 3: Substitute the value:

To find the residue of the function f(z)  at  $z=\infty$ substitute  $\frac{1}{Z}$ for  z and then you have,

 $f\left(\frac{1}{z}\right)=a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+...+\frac{a_r}{z^r}+b_{1}z+b_2{z}^2+b_3{z}^3+...$

Since, $f\left(\frac{1}{z}\right)$  has pole of order r  at,  z = 0.

So, $z= \infty$   is singular point of  f(z).

 

By residue theorem as follows:

 ${\oint }_{c}f\left(z\right)dz=2\mathrm{\pi i}\left(\mathrm{Resf}\left(\infty \right)\right)$

Here,  $\mathrm{Resf}\left(\infty \right)$  is the residue of  f(z) at $z=\infty$  .

Now, 

 $z=\frac{1}{Z}$

$dz=-\frac{1}{z^2}dZ$                                                                                                               ….. (1)

Substitute the values from equations as follows:

 $$\begin{gathered} \oint f\left(z\right)dz={\int }_c\frac{1}{z^2}f\left(\frac{1}{z}\right)dz \\ =-2\pi i\left(Res\left(\frac{1}{z^2}f{\left(\frac{1}{z}\right)}_{z=\infty }dz\right)\right) \end{gathered}$$

Step 4: Substitute the value further

The direction of contour integration changes since the mapping $z=\frac{1}{Z}$  transforms the exterior region of circle  C. $\left|z\right|=M$  into interior region of circle $\left|z\right|=\frac{1}{M}$ .

 $\frac{1}{z^2}f\left(\frac{1}{z}\right)=\frac{a_0}{Z^2}+\frac{a_1}{Z^3}+\frac{a_2}{Z^4}+...+\frac{a_r}{Z^{r+2}}+\frac{b_1}{Z}+b_2+b_{3}z+...$

 

 

Since the residue of $\frac{1}{z^2}f\left(\frac{1}{z}\right)$  at z = 0  is the coefficient of  $\frac{1}{z}$.

Therefore, 

 $$\begin{gathered} {\oint }_{c}f\left(z\right)={\oint }_c-\frac{1}{z^2}f\left(\frac{1}{z}\right)dz \\ =-2\pi i\left(-b_1\right) \end{gathered}$$

 

Here, the negative sign of the residue is due to the change   the direction of contour integration.

 

Hence, it is proved that the result of integrating Laurent series about infinity in the positive (contour clockwise) direction is  $-2\pi i\left(-b_1\right)$.