For each of the following functions find the first few terms of each of the Laurent series about the origin, that is, one series for each annular ring between singular points. Find the residue of each function at the origin. (Warning: To find the residue, you must use the Laurent series, which converges near the origin.) Hints: See Problem 2. Use partial fractions as in equations (4.5) and (4.7). Expand a term $\frac{\mathbf{1}}{\left(\mathbf{z}\mathbf{-}\mathbf{\alpha }\right)}$ in powers of z to get a series convergent for $\left|\mathbf{z}\right|\mathbf{<}\mathbf{\alpha }$, and in powers of $\frac{\mathbf{1}}{\mathbf{z}}$ to get a series convergent for $\left|\mathbf{z}\right|\mathbf{>}\mathbf{\alpha }$.

$\mathbf{f}\left(\mathbf{z}\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{z}\left(\mathbf{z}\mathbf{-}\mathbf{1}\right)\left(\mathbf{z}\mathbf{-}\mathbf{2}\right)}$

The given function is$\mathrm{f}\left(\mathrm{z}\right)=\frac{1}{\mathrm{z}\left(\mathrm{z}-1\right)\left(\mathrm{z}-2\right)}$

Laurent’s series, additionally referred to as Laurent’s growth, of a posh perform f(z) is outlined as an illustration of that perform in terms of series that features the terms of negative degree.

Use Laurent’s theorem to expand f(z)

$\begin{array}{c}f\left(z\right)={\alpha }_0+{\alpha }_{1}z-z_0+{\alpha }_2{\left(z-z_0\right)}^2+...+{\alpha }_n{\left(z-z_0\right)}^n+{\alpha }_{n+1}{\left(z-z_0\right)}^{n+1}\\ +{\alpha }_{n+2}{\left(z-z_0\right)}^{n+2}+...+\frac{b_1}{z-z_0}+\frac{b_2}{{\left(z-z_0\right)}^2}+..\end{array}$

Consider the function:

$f\left(z\right)=\frac{1}{z\left(z-1\right)\left(z-2\right)}$

This function has three singular points at $z=0,1,2$.

So there are two circles $C_1\&C_2$about z=0, and Laurent’s series about z=0, one series valid in each of the three regions $R_1\left(0<\left|z\right|<1\right),R_2\left(1<\left|z\right|<2\right)\mathrm{and}{\mathrm{R}}_3\left(\left|\mathrm{z}\right|>2\right)$

Rewrite the function as:

$f\left(z\right)=\frac{1}{z}\left(\frac{1}{z-2}-\frac{1}{z-1}\right)$

Consider the figure shown below:

Take the region $0<\left|z\right|<1:$

$\begin{array}{l}\left(z\right)=\frac{1}{z}\left[\frac{1}{-2}{\left(1-\frac{z}{2}\right)}^{-1}+{\left(1-z\right)}^{-1}\right]\\ =\frac{1}{z}\left[-\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+...\right)+\left(1+z+z^2+...\right)\right]\\ =\left(-\frac{1}{2z}-\frac{1}{4}-\frac{z}{8}-\frac{z^2}{16}-...\right)+\left(\frac{1}{z}+1+z+z^2+...\right)\\ =\frac{3}{4}+\frac{1}{2z}+\frac{7z}{8}+\frac{15}{16}{z}^2+...\end{array}$

Also, determine the value of the residue

$R\left(0\right)=\frac{1}{2}$

Take the region $1<\left|z\right|<2$:

$\begin{array}{l}f\left(z\right)=\frac{1}{z}\left(\frac{1}{z-2}-\frac{1}{z-1}\right)\\ =\frac{1}{z}\left[-\frac{1}{2}{\left(1-\frac{z}{2}\right)}^{-1}-\frac{1}{z}{\left(1-\frac{1}{z}\right)}^{-1}\right]\\ =-\frac{1}{2z}\left(1+\frac{z}{2}+\frac{z^2}{2^2}+\frac{z^3}{2^3}+...\right)-\frac{1}{z^2}\left(1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...\right)\\ =\left(-\frac{1}{2z}-\frac{1}{4}-\frac{z}{2^3}-\frac{z^2}{2^4}-...\right)-\left(\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+\frac{1}{z^5}+...\right)\end{array}$

Continue further as shown below

$\begin{array}{l}f\left(z\right)=\left(-\frac{1}{4}-\frac{z}{2^3}-\frac{z^2}{2^4}-...\right)-\left(\frac{1}{2z}+\frac{1}{z^2}+\frac{1}{z^3}+\frac{1}{z^4}+...\right)\\ =...-z^4-z^3-z^{-2}-\frac{1}{2}{z}^{-1}-\frac{1}{2^2}-\frac{z}{2^3}-\frac{z^2}{z^4}-...\end{array}$

Take the region $\left|z\right|>2$:

$\begin{array}{l}f\left(z\right)=\frac{1}{z}\left[\frac{1}{z}{\left(1-\frac{2}{z}\right)}^{-1}-\frac{1}{z}{\left(1-\frac{1}{z}\right)}^{-1}\right]\\ =\frac{1}{z^2}\left(1+\frac{2}{z}+\frac{2^2}{z^2}+\frac{2^3}{z^3}+...\right)-\frac{1}{z^2}\left(1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...\right)\\ =\left(\frac{1}{z^2}+\frac{2}{z^3}+\frac{2^2}{z^4}+\frac{2^3}{z^5}+...+\right)\left(-\frac{1}{z^2}-\frac{1}{z^3}-\frac{1}{z^4}-\frac{1}{z^5}-...\right)\\ =\frac{1}{z^3}+\frac{3}{z^4}+\frac{7}{z^5}+...\end{array}$

Therefore, the function is:

$f\left(z\right)=z^{-3}+3z^{-4}+7z^{-5}+...$

Hence, the final equation for the function and value of the residue is:

role="math" localid="1664349419414" $$\begin{gathered} f\left(z\right)=\left\{\begin{array}{l}\frac{3}{4}+\frac{1}{2z}+\frac{7z}{8}+\frac{15}{16}{z}^2+...;\left(0<\left|z\right|<1\right)\\ ...-z^{-4}-z^{-3}-z^{-2}-\frac{1}{2}{z}^{-1}-\frac{1}{2^2}-\frac{z}{2^3}-\frac{z^2}{2^4}-...\left(1-<\left|z\right|<2\right)\\ z^{-3}+3z^{-4}+7z^{-5}+...;\left(\left|z\right|>2\right)\end{array}\right. \\ R\left(0\right)=\frac{1}{2} \end{gathered}$$