What is the taylor and Laurent series of this?

>>9090144
[eqn] z + 2 = \frac{5}{2}(z - 1) - \frac{3}{2} (z-3) [/eqn]
so
[eqn] w = \frac{3}{2} \frac{1}{1-z} - \frac{5}{6} \frac{1}{1 - \frac{z}{3}} \\
= \frac{3}{2} \sum_{k=0}^\infty z^k - \frac{5}{6} \sum_{k=0}^\infty \left( \frac{z}{3} \right)^k \\
= \sum_{k=0}^\infty \left(\frac{3}{2} - \frac{5}{6} \frac{1}{3^k} \right) z^k[/eqn]

>>9090182
this

Taylor series involves derivatives, but when it comes to complex valued functions, derivatives only "make sense" for holomorphic functions.

Holomorphic is "equivalent" to analytic, which means the function looks like a power series. When Laurent series is mentioned for complex functions, there could be negative exponent terms, i.e.

[math]f(z) = \sum_{n = -\infty}^{\infty}a_n(z-z_0)^n[/math]

If there are no negative exponent terms, then f(z) is holomorphic on the disk centered at z_0 with radius determined by the radius of convergence. Similarly, a laurent series will be holomorphic on an annulus

>>9090182
How you got 5/2 for (z-1) and -3/2 for (z-3)?

>>9090182
Thanks a lot, but for taylor, I still don't know how to do it, because you did partial fractions for laurent

PFD makes the Tailor series ezpz.

>>9090144
Taylor series use non-negative exponents in the power series.

Laurent series can use negative exponents.

Usually you can use L'hopital's rule to figure out the residues at a singularity.