Question

obtain all the lawent series of the function 7Z-2/(z+1)z(z-2) aboutZ0=-1​

Answer 1

Answer:

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Look for the singular points of the function. They are at z=−1 , z=0 and z=2 . The curve |z+1|=3 is a circle with radius 3 centred at z=−1 and the curve |z+1|=1 is a circle with radius 1 centred at z=−1 . So the region is between these two circles and the function has no singular points in the region, but the centre is a singularity. There is also a singular point at z=0 on the inner circle and another at z=2 on the outer circle. If you use partial fractions, the function is of the form Az+Bz+1+Cz−2 . The term Bz+1 is the first term of the Laurent series. Expand the other terms by Taylor’s theorem about z=−1 (or just use the geometric series) and sum coefficients of like powers. The first has radius of convergence 1 and the second has radius of convergence 2 . So they both converge on the required region

I’ll leave the details to you—I’m sure you’re capable.