Consider f ( z ) = z − 1 z 2 − 2 z − 3 = z − 1 ( z − 3 ) ( z + 1 ) f(z)=\frac{z-1}{z^{2}-2 z-3}=\frac{z-1}{(z-3)(z+1)} f ( z ) = z 2 − 2 z − 3 z − 1 = ( z − 3 ) ( z + 1 ) z − 1
Using partial fraction we get
z − 1 ( z − 3 ) ( z + 1 ) = a z − 3 + b z + 1 \frac{z-1}{(z-3)(z+1)}=\frac{a}{z-3}+\frac{b}{z+1} ( z − 3 ) ( z + 1 ) z − 1 = z − 3 a + z + 1 b
i.e. z-1=a(z+1)+b(z-3)
put z=3 , we get a = 1 2 a=\frac{1}{2} a = 2 1
put z=-1 , we get b = 1 2 b=\frac{1}{2} b = 2 1
z − 1 ( z − 3 ) ( z + 1 ) = 1 2 z − 3 + 1 2 z + 1 \frac{z-1}{(z-3)(z+1)}=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} ( z − 3 ) ( z + 1 ) z − 1 = z − 3 2 1 + z + 1 2 1
Hence f(z) is not analytic at z=-1 and z=3
∴ \therefore ∴
∗ ∣ z ∣ < 1 ∗ 1 < ∣ z ∣ < 3 ∗ ∣ z ∣ > 3 * |z|<1
\\
* 1<|z|<3
\\
* |z|>3 ∗ ∣ z ∣ < 1 ∗ 1 < ∣ z ∣ < 3 ∗ ∣ z ∣ > 3
(i)
∣ z ∣ < 1 ∵ ∣ z ∣ < 1 , ∣ z ∣ < 3 |z|<1
\\\because|z|<1,|z|<3 ∣ z ∣ < 1 ∵ ∣ z ∣ < 1 , ∣ z ∣ < 3
∴ f ( z ) = 1 2 z − 3 + 1 2 z + 1 = 1 2 ⋅ 1 − 3 ⋅ 1 1 − z 3 + 1 2 1 1 + z = − 1 6 ( 1 − z 3 ) − 1 + 1 2 ( 1 + z ) − 1 \begin{aligned}
&\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\
&=\frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2} \frac{1}{1+z} \\
&=-\frac{1}{6}\left(1-\frac{z}{3}\right)^{-1}+\frac{1}{2}(1+z)^{-1}
\end{aligned} ∴ f ( z ) = z − 3 2 1 + z + 1 2 1 = 2 1 ⋅ − 3 1 ⋅ 1 − 3 z 1 + 2 1 1 + z 1 = − 6 1 ( 1 − 3 z ) − 1 + 2 1 ( 1 + z ) − 1
= − 1 6 ( 1 + z 3 + ( z 3 ) 2 + ( z 3 ) 3 + … … ) + 1 2 ( 1 − z + z 2 − z 3 + … … ) f ( z ) = 1 3 − 5 9 z + 13 27 z 2 − … … \begin{aligned}
&=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots \ldots\right)+\frac{1}{2}\left(1-z+z^{2}-z^{3}+\ldots \ldots\right) \\
&f(z)=\frac{1}{3}-\frac{5}{9} z+\frac{13}{27} z^{2}-\ldots \ldots
\end{aligned} = − 6 1 ( 1 + 3 z + ( 3 z ) 2 + ( 3 z ) 3 + …… ) + 2 1 ( 1 − z + z 2 − z 3 + …… ) f ( z ) = 3 1 − 9 5 z + 27 13 z 2 − ……
Above is required Taylor's series
(ii) 1<|z|<3
We get ∣ 1 z ∣ < 1 \left|\frac{1}{z}\right|<1 ∣ ∣ z 1 ∣ ∣ < 1 ∣ z 3 ∣ < 1 \left|\frac{z}{3}\right|<1 ∣ ∣ 3 z ∣ ∣ < 1
∴ f ( z ) = 1 2 z − 3 + 1 2 z + 1 = 1 2 ⋅ 1 − 3 ⋅ 1 1 − z 3 + 1 2 ⋅ 1 z ⋅ 1 1 + 1 z = 1 − 6 ⋅ 1 1 − z 3 + 1 2 z ⋅ 1 1 + 1 z = − 1 6 ( 1 − z 3 ) − 1 + 1 2 z ( 1 + 1 z ) − 1 = − 1 6 ( 1 + z 3 + ( z 3 ) 2 + ( z 3 ) 3 + … … ) + 1 2 z ( 1 − ( 1 z ) + ( 1 z ) 2 − ( 1 z ) 3 + … ) 2 + 1 2 ( 1 z − ( 1 z ) 2 + ( 1 z ) 3 − ( 1 z ) 4 + … f ( z ) = − 1 6 ( 1 + z 3 + ( z 3 ) 2 + ( z 3 ) 3 + … ) \begin{aligned}
&\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\
&=\frac{1}{2} \cdot \frac{1}{-3} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2} \cdot \frac{1}{z} \cdot \frac{1}{1+\frac{1}{z}} \\
&=\frac{1}{-6} \cdot \frac{1}{1-\frac{z}{3}}+\frac{1}{2 z} \cdot \frac{1}{1+\frac{1}{z}} \\
&=-\frac{1}{6}\left(1-\frac{z}{3}\right)^{-1}+\frac{1}{2 z}\left(1+\frac{1}{z}\right)^{-1} \\
&=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots \ldots\right)+\frac{1}{2 z}\left(1-\left(\frac{1}{z}\right)+\left(\frac{1}{z}\right)^{2}-\left(\frac{1}{z}\right)^{3}+\ldots\right)^{2}+\frac{1}{2}\left(\frac{1}{z}-\left(\frac{1}{z}\right)^{2}+\left(\frac{1}{z}\right)^{3}-\left(\frac{1}{z}\right)^{4}+\ldots\right. \\
&f(z)=-\frac{1}{6}\left(1+\frac{z}{3}+\left(\frac{z}{3}\right)^{2}+\left(\frac{z}{3}\right)^{3}+\ldots\right)
\end{aligned} ∴ f ( z ) = z − 3 2 1 + z + 1 2 1 = 2 1 ⋅ − 3 1 ⋅ 1 − 3 z 1 + 2 1 ⋅ z 1 ⋅ 1 + z 1 1 = − 6 1 ⋅ 1 − 3 z 1 + 2 z 1 ⋅ 1 + z 1 1 = − 6 1 ( 1 − 3 z ) − 1 + 2 z 1 ( 1 + z 1 ) − 1 = − 6 1 ( 1 + 3 z + ( 3 z ) 2 + ( 3 z ) 3 + …… ) + 2 z 1 ( 1 − ( z 1 ) + ( z 1 ) 2 − ( z 1 ) 3 + … ) 2 + 2 1 ( z 1 − ( z 1 ) 2 + ( z 1 ) 3 − ( z 1 ) 4 + … f ( z ) = − 6 1 ( 1 + 3 z + ( 3 z ) 2 + ( 3 z ) 3 + … )
This is required Laurent's series
(iii) |z|>3
∵ ∣ z ∣ > 3 , clearly, ∣ z ∣ > 1 ∴ ∣ z 3 ∣ > 1 and ∣ z ∣ > 1 i.e. ∣ 3 z ∣ < 1 and ∣ 1 z ∣ < 1 ∴ f ( z ) = 1 2 z − 3 + 1 2 z + 1 = 1 2 z ⋅ 1 1 − 3 z + 1 2 z ⋅ 1 1 + 1 z = 1 2 z ( 1 − 3 z ) − 1 + 1 2 z ( 1 + 1 z ) − 1 = 1 2 z ( 1 + ( 3 z ) + ( 3 z ) 2 + ( 3 z ) 3 + … ) + 1 2 z ( 1 − ( 1 z ) + ( 1 z ) 2 − ( 1 z ) 3 + … . ) = 1 z + 1 z 2 + 5 z 3 + 13 z 4 + … ⋯ ⋯ f ( z ) = 1 z + 1 z 2 + 5 z 3 + 13 z 4 + … … … . \begin{aligned}
&\because|z|>3, \text { clearly, }|z|>1 \\
&\therefore\left|\frac{z}{3}\right|>1 \text { and }|z|>1 \\
&\text { i.e. }\left|\frac{3}{z}\right|<1 \text { and }\left|\frac{1}{z}\right|<1 \\
&\therefore f(z)=\frac{\frac{1}{2}}{z-3}+\frac{\frac{1}{2}}{z+1} \\
&=\frac{1}{2 z} \cdot \frac{1}{1-\frac{3}{z}}+\frac{1}{2 z} \cdot \frac{1}{1+\frac{1}{z}} \\
&=\frac{1}{2 z}\left(1-\frac{3}{z}\right)^{-1}+\frac{1}{2 z}\left(1+\frac{1}{z}\right)^{-1} \\
&=\frac{1}{2 z}\left(1+\left(\frac{3}{z}\right)+\left(\frac{3}{z}\right)^{2}+\left(\frac{3}{z}\right)^{3}+\ldots\right)+\frac{1}{2 z}\left(1-\left(\frac{1}{z}\right)+\left(\frac{1}{z}\right)^{2}-\left(\frac{1}{z}\right)^{3}+\ldots .\right) \\
&=\frac{1}{z}+\frac{1}{z^{2}}+\frac{5}{z^{3}}+\frac{13}{z^{4}}+\ldots \cdots \cdots \\
&f(z)=\frac{1}{z}+\frac{1}{z^{2}}+\frac{5}{z^{3}}+\frac{13}{z^{4}}+\ldots \ldots \ldots .
\end{aligned} ∵ ∣ z ∣ > 3 , clearly, ∣ z ∣ > 1 ∴ ∣ ∣ 3 z ∣ ∣ > 1 and ∣ z ∣ > 1 i.e. ∣ ∣ z 3 ∣ ∣ < 1 and ∣ ∣ z 1 ∣ ∣ < 1 ∴ f ( z ) = z − 3 2 1 + z + 1 2 1 = 2 z 1 ⋅ 1 − z 3 1 + 2 z 1 ⋅ 1 + z 1 1 = 2 z 1 ( 1 − z 3 ) − 1 + 2 z 1 ( 1 + z 1 ) − 1 = 2 z 1 ( 1 + ( z 3 ) + ( z 3 ) 2 + ( z 3 ) 3 + … ) + 2 z 1 ( 1 − ( z 1 ) + ( z 1 ) 2 − ( z 1 ) 3 + … . ) = z 1 + z 2 1 + z 3 5 + z 4 13 + …⋯⋯ f ( z ) = z 1 + z 2 1 + z 3 5 + z 4 13 + ……… .
This is the required Laurent's series
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