That series doesn't converge for |z|<1.squenshl said:I let t = z
So 1/(t+i) = 1/t(1+i/t) which is a geometric series
= 1/t(1-(i/t)+(-1/t)^2-...
= 1/t - i/t2 + i2/t3 + ...
= 1/z - i/z2 - 1/z3 + ...
You don't. The 1/z and 1/z2 terms are already in the correct form.squenshl said:How do I find the geometric series for 1/z and -i/z2
A Laurent series is a representation of a complex function as an infinite sum of terms, where each term is a power of the complex variable z. The series can include both positive and negative powers of z.
To calculate a Laurent series, you can use the formula 1/(z2(z+i)) = 1/(z2) - 1/(z2+i) and then expand each term using the geometric series formula. This will give you an infinite sum of powers of z, with coefficients determined by the original function.
Laurent series are useful in complex analysis as they allow us to approximate complex functions and better understand their behavior near singularities, such as poles or branch points. They also help us to solve complex integration problems and find solutions to differential equations.
If the function has a singularity at z=0, then the Laurent series will have a negative power term in addition to positive power terms. This is because the function cannot be expanded as a Taylor series around z=0 due to the singularity, so the Laurent series includes both positive and negative powers to account for the singularity.
No, not all complex functions can be represented by a Laurent series. The function must be analytic in the region around the point of expansion, meaning it must be continuous and have a derivative at that point. Functions with essential singularities, such as e1/z, cannot be represented by a Laurent series.