Expand into Laurent's Series

Question

Expand into Laurent's series after the power of z-a for the following function

$$ f(z)=\frac{z+z^2}{(1-z)^3} $$for a = 0 .

I know the formula for the Laurent's series but I don't know how to apply it.

Answer 1

Hint: For $|z| <1$ we have $$(1-z)^{-3}=1+\frac {-3} 1 (-z)+\frac {(-3) (-4)} {(1)(2)} (-z)^{2}$$ $$+\frac {(-3) (-4)(-5)} {(1)(2)(3)} (-z)^{3}+...$$ Now multiply by $z+z^{2}$

Answer 2

I would start from the expansion of $\dfrac1{1-z}$ to obtain the expansion of $\dfrac 1{(1-z)^3}$: $$\dfrac 1{(1-z)^3}=\frac12\biggl(\frac1{1-z}\biggr)\!\vphantom{\biggl)}''=\frac12\sum_{n=1}^{\infty}n(n-1)z^{n-2}=\sum_{n=1}^{\infty}\binom n2z^{n-2}$$ (with the convention that $\binom 12=0\,$).

Now multiply by $z$ and $z^2$: $$ \frac{z+z^2}{(1-z)^3} =\sum_{n=1}^{\infty}\binom n2z^{n-1}+\sum_{n=1}^{\infty}\binom n2z^{n}=\sum_{n=1}^{\infty}\biggl[\binom n2+\binom{n+1}2\biggr]z^{n} $$ and simplify the coefficients.