# Why do these laurent series approaches conflict?

I was working on a problem of finding the Laurent series of $$\frac{1}{z-3}$$ that converges where $$|z-4| > 1$$

So I had one approach, let $$u=z-4$$ then:

$$\frac{1}{z-3} = \frac{1}{1+u}$$

$$= \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$

$$= \frac{1}{z-4} - \frac{1}{(z-4)^2} + ...$$

But this apparently incorrect.

The correct answer is found by noting:

$$\frac{1}{z-3} = \frac{1}{z-4 + 1} = \frac{1}{z-4} \frac{1}{1 - \frac{-1}{z-4}} = -\frac{1}{(z-4)^2} + ...$$

Where did I go wrong?

• Your first one is correct. $\frac {1}{z-3} = \frac {1}{z-4} - \frac {1}{(z-4)^2} + \cdots$ Try evaluating $z = 6\cdots \frac {1}{3} = \frac 12 - \frac 14 + \frac 18-\cdots$ – Doug M Apr 5 at 1:59
• The Princeton review book says the first term is of order 2, and they make... you know what I think they are plain wrong here. Since their factorization introduces a first order term in either expansion direction – frogeyedpeas Apr 5 at 2:00

$${1\over 1+x} = 1 - x + x^2 - x^3 + \cdots$$ when $$|x| < 1$$.
This is derived from $${1 \over 1-x} = 1 + x + x^2 + \cdots$$ with $$|x| < 1$$.
• I’m not sure I understand. I’m using the formula for $|x|>1$ – frogeyedpeas Apr 5 at 2:03
• @frogeyedpeas I believe that is the expansion of $\frac{1}{1+ {1\over u}}$ – U2647 Apr 5 at 2:11
• You can also use this formula, but remember to divide $u$ on both sides. – U2647 Apr 5 at 2:12
• It’s one of 2 expansions. $\frac{1}{1+x} = \frac{1}{x}-\frac{1}{x^2} + ...$ – frogeyedpeas Apr 5 at 2:12
\begin{align} \frac1{z-4}\frac1{1+\frac1{z-4}} &=\frac1{z-4}\left(1-\frac1{z-4}+\frac1{(z-4)^2}-\dots\right)\\ &=\frac1{z-4}-\frac1{(z-4)^2}+\frac1{(z-4)^3}-\dots \end{align} The series starts with $$\frac1{z-4}$$, not $$\frac1{(z-4)^2}$$.