Why do these laurent series approaches conflict?

Question

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?

Answer 1

${1\over 1+x} = 1 - x + x^2 - x^3 + \cdots$ when $|x| < 1$.

This is derived from ${1 \over 1-x} = 1 + x + x^2 + \cdots$ with $|x| < 1$.

Answer 2

$$ \begin{align} \frac1{z-4}\frac1{1+\frac1{z-4}} &=\frac1{z-4}\left(1-\frac1{z-4}+\frac1{(z-4)^2}-\dots\right)\\ &=\frac1{z-4}-\frac1{(z-4)^2}+\frac1{(z-4)^3}-\dots \end{align} $$ The series starts with $\frac1{z-4}$, not $\frac1{(z-4)^2}$.