Find all possible Taylor and Laurent series expansions for the function(2z + 3)/((z + 1)(z + 2)) around z = 1 and specify the domain of validity.

The function can be written as a partial fraction:

(2z + 3)/((z + 1)(z + 2)) = A/(z + 1) + B/(z + 2)

Solving for A and B gives A = -1 and B = 3. So the function can be rewritten as:

-1/(z + 1) + 3/(z + 2)

Now, we can find the Taylor series expansion around z = 1 for each term separately.

For -1/(z + 1), we can write it as -1/(1 + (z - 1)), and then use the geometric series formula to get the Taylor series:

-1/(1 + (z - 1)) = -1 - (z - 1) - (z - 1)^2 - (z - 1)^3 - ...

For 3/(z + 2), we can write it as 3/(2 + (z - 2)), and then use the geometric series formula to get the Taylor series:

3/(2 + (z - 2)) = 3/2 - 3/4*(z - 2) + 9/16*(z - 2)^2 - 27/64*(z - 2)^3 + ...

Adding these two series together gives the Taylor series for the original function:

-1 - (z - 1) - (z - 1)^2 - (z - 1)^3 - ... + 3/2 - 3/4*(z - 2) + 9/16*(z - 2)^2 - 27/64*(z - 2)^3 + ...

This series is valid for |z - 1| < 1 and |z - 2| < 2.

For the Laurent series, we can use the same partial fractions, but now we need to find the series around z = 1. This means we need to express each term as a series in (z - 1).

For -1/(z + 1), we can write it as -1/(2 - (1 - z)), and then use the geometric series formula to get the Laurent series:

-1/(2 - (1 - z)) = -1/2 + 1/4*(1 - z) - 1/8*(1 - z)^2 + 1/16*(1 - z)^3 - ...

For 3/(z + 2), we can write it as 3/(3 - (1 - z)), and then use the geometric series formula to get the Laurent series:

3/(3 - (1 - z)) = 1 - 1/3*(1 - z) + 1/9*(1 - z)^2 - 1/27*(1 - z)^3 + ...

Adding these two series together gives the Laurent series for the original function:

-1/2 + 1/4*(1 - z) - 1/8*(1 - z)^2 + 1/16*(1 - z)^3 - ... + 1 - 1/3*(1 - z) + 1/9*(1 - z)^2 - 1/27*(1 - z)^3 + ...

This series is valid for 1 < |z - 1| < 2.