Question

F(z):=1 / ((z+1)(z-3)) by a Laurent series on the annuli A:=z.

Answer 1

To find the Laurent series of f(z) = 1/((z+1)(z-3)) on the annulus A:=z ∈ ℂ: 1 < , we can use partial fraction decomposition and equate the coefficients of the like powers of z.

Step-by-step explanation:

The function f(z) is given by f(z) = 1/((z+1)(z-3)). To find the Laurent series of f(z) on the annulus A:=z, we can use partial fraction decomposition.

First, we decompose f(z) into partial fractions: f(z) = A/(z+1) + B/(z-3), where A and B are constants.

To find A and B, we can multiply both sides of the equation by (z+1)(z-3) and equate the coefficients of the like powers of z. Solving for A and B will give us the values of the constants.

Once A and B are found, we can express f(z) as a Laurent series: f(z) = ∑(A_n/(z-a)^n), where A_n is the coefficient of (z-a)^n for each term in the series.