To find the Laurent series of the function f(z) = 4z+3/z(z-3)(z+2) in the region mod z = 1, we need to find the poles of the function within the region.
The function has poles at z=0, z=3, and z=-2. Since the region is mod z = 1, the only pole that lies within the region is z=0.
Now, we can write the Laurent series of f(z) about z=0 as:
f(z) = a_{-2}(z-0)^{-2} + a_{-1}(z-0)^{-1} + a_0 + a_1(z-0) + a_2(z-0)^2 + ...
To find the coefficients of the Laurent series, we can use the formula:
a_k = (1/2πi) ∮(f(z)(z-0)^{k+1})dz
where the contour of integration is a circle of radius 1 centered at z=0.
The coefficient a_{-2} can be found as:
a_{-2} = (1/2πi) ∮(f(z)/z^2)dz
We can use partial fractions to write f(z)/z^2 as:
f(z)/z^2 = 1/z - 3/(z-3) + 2/(z+2)
Using the formula for each term, we get:
a_{-2} = Res[f(z)/z^2, z=0] = 0
Similarly, we can find the coefficients a_{-1}, a_0, a_1, and a_2 as:
a_{-1} = Res[f(z)/z, z=0] = 4
a_0 = f(0) = 3/(-3)(-2) = -1/2
a_1 = f'(0) = -9/(-3)(-2) = 3/2
a_2 = (1/2πi) ∮(f(z)z)dz = Res[f(z), z=0] = 0
Therefore, the Laurent series of f(z) about z=0 in the region mod z=1 is:
f(z) = 4/z - 1/2 + 3/2 z + O(z^2)