A question about Laurent series.

Question

$p(z)=a^2z+bz+c$, how to get the Laurent series of $\frac{p(z)}{z-1}$ in $\mathbb C\setminus\{1\}$? ($a$ is not equal to zero)

Answer 1

You can guess that $p(z) = a(z-1)^2 + d(z-1) + e = az^2 + bz + c,$ and solve for $d,e$

You can find $p(z+1)$ and that will point you to the correct coefficients, too.

Or, you can find the quotient, $\frac {p(z)}{z-1} = q(z) + \frac {r}{z-1}$ and take the remainder.

Then divide the quotient of $\frac {q(z)}{z-1}$ and take the remainder.

and the last quotient is the constant $a.$

These will be the coefficients of the $(z-1)^0, (z-1)^1, (z-1)^2$ terms, respectively

Answer 2

Without solving any equation, you can find the coefficients in the expansion of $p(z)$ by powers of $z-1$ by the method of successive divisions by $z-1$, exactly the same way as you find the expansion of a natural number in basis $b$ by successive divisions of its decimal expansion by the basis.

You should find $$p(z)=(a+b+c) +(2a+b)(z-1)+a(z-1)^2,$$ so that $$\frac{p(z)}{z-1}=\frac{a+b+c}{z-1} +(2a+b)+a(z-1).$$