# Laurent Series From Cracking the GRE (Mistake?)

The problem asks to find the Laurent series for $f(z)=\frac{1}{z-3}$ in the annulus $|z-4|>1$. I found the answer to be $\sum\limits_{n=0}^{\infty}(-1)^n(z-4)^{-n-1}$. However, the book states that the answer is $\sum\limits_{n=1}^{\infty}(-1)^n(z-4)^{-n-1}$. I think this is wrong, since the first term in the expansion should be $\frac{1}{z-4}$. Is the book wrong, or did I do something incorrect?

• I saw this as well, it is a mistake – operatorerror Sep 10 '16 at 21:13
• I thought so. Thanks! – user269711 Sep 10 '16 at 21:25

$$\frac1{z-3}=\frac1{1+(z-4)}=\frac1{z-4}\cdot\frac1{1+\frac1{z-4}}=\frac1{z-4}\sum_{n=0}^\infty(-1)^n(z-4)^{-n}=$$
$$=\sum_{n=0}^\infty(-1)^n(z-4)^{-n-1}$$