In the Laurent series expression of $f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}$ valid for 0 < |z - 1|< 1, the co-efficient of $\frac{1}{\left(z-1\right)}$ is

Understanding the Laurent Series Coefficient Problem

The question asks us to find the coefficient of the term $\frac{1}{z-1}$ in the Laurent series expansion of the function $f\left(z\right)=\frac{1}{\left(z-1\right)\left(z-2\right)}$. The expansion needs to be valid within the specific annulus region defined by $0<|z-1|<1$.

Step 1: Partial Fraction Decomposition

First, we decompose the given function $f(z)$ into partial fractions. Let:

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

To find the constants A and B, we multiply both sides by $\left(z-1\right)\left(z-2\right)$:

$$1=A\left(z-2\right)+B\left(z-1\right)$$

Now, we substitute specific values of z to solve for A and B:

- Let $z=1$: $1=A(1-2)+B(1-1)⟹1=A(-1)⟹A=-1$
- Let $z=2$: $1=A(2-2)+B(2-1)⟹1=B(1)⟹B=1$

So, the partial fraction decomposition is:

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

Step 2: Laurent Series Expansion

We need the Laurent series expansion around $z=1$ in the region $0<|z-1|<1$. Let's make the substitution $w=z-1$. This implies $z=w+1$. The given region becomes $0<|w|<1$.

The function $f(z)$ can be rewritten in terms of $w$:

$$f(z)=\frac{-1}{w}+\frac{1}{(w+1)-2}=\frac{-1}{w}+\frac{1}{w-1}$$

The first term, $\frac{-1}{w}$, is already in the form required for the Laurent series (a term with $(z-1{)}^{-1}$). We need to expand the second term, $\frac{1}{w-1}$, as a power series in $w$ that is valid for $0<|w|<1$.

Let's rewrite the second term:

$$\frac{1}{w-1}=-\frac{1}{1-w}$$

Since $|w|<1$ in our region, we can use the geometric series formula $\frac{1}{1-x}={\sum }_{n=0}^{\infty }{x}^n$ for $|x|<1$. Applying this with $x=w$:

$$-\frac{1}{1-w}=-\sum _{n=0}^{\infty }{w}^n=-(1+w+w^2+w^3+...)$$

Now, substitute this back into the expression for $f(z)$:

$$f(z)=\frac{-1}{w}-\sum _{n=0}^{\infty }{w}^n$$

$$f(z)=\frac{-1}{w}-(1+w+w^2+w^3+...)$$

Finally, substituting back $w=z-1$:

$$f(z)=\frac{-1}{z-1}-\left(1+(z-1)+(z-1{)}^2+(z-1{)}^3+...\right)$$

This is the Laurent series expansion of $f(z)$ around $z=1$ valid for $0<|z-1|<1$.

Step 3: Identifying the Coefficient

The Laurent series is given by ${\sum }_{n=-\infty }^{\infty }{a}_n(z-c{)}^n$. In our case, $c=1$. The series we found is:

$$f(z)=(-1)\cdot (z-1{)}^{-1}-1\cdot (z-1{)}^0-1\cdot (z-1{)}^1-1\cdot (z-1{)}^2-...$$

The question specifically asks for the coefficient of $\frac{1}{z-1}$, which corresponds to the coefficient $a_{-1}$ where the power of $(z-1)$ is -1.

Looking at the expansion, the term with $(z-1{)}^{-1}$ is $\frac{-1}{z-1}$. Therefore, the coefficient $a_{-1}$ is -1.