Binomial Expansion

[lilshai]

Hello...
I have this problem that says:
Find the Taylor series of f(z) = 3/(3z - z²)) around z = 1.

I know we have to start by writing the partial fraction expansion out, so we have

f(z) = 3/(3z - z²)) = 1/z + 1/(3-z) but after that I get stuck. The solution says to use the Binomial Expansion and I'm not sure why and how to use it.

Here is the rest of the solution:

f(z) = 1/(1 + (z-1)) + 1/(2 - (z-1)) --->> Expand around z = 1
= 1 - (z-1) + (z-1)² + ... + (-1)ⁿ(z-1)ⁿ+...+ 1/2 + (z-1)/(2²) + (z-1)²/2³ + ... + (z-1)ⁿ/2ⁿ⁺¹+...

= 3/2 - 3/4(z-1) + 9/8(z-1)² - ... + [1/(2ⁿ⁺¹) + (-1)ⁿ](z-1)ⁿ + ...

Can anyone help with this?

 

[galactus]

The Taylor expansion for 1/z, about x=1 is:

\(\displaystyle \L\\1-(z-1)+(z-1)^{2}-(z-1)^{3}+(z-1)^{4}-........\)

The Taylor for \(\displaystyle \L\\\frac{1}{3-z}\), about z=1 is:

\(\displaystyle \L\\\frac{1}{2}+\frac{1}{4}(z-1)+\frac{1}{8}(z-1)^{2}+\frac{1}{16}(z-1)^{3}+\frac{1}{32}(z-1)^{4}+.............\)

Add them up.