1 Given the complex number $m$ satisfies $m+\frac{1}{m}=1$, then $m^{2008}+\frac{1}{m^{2009}}=$ $\qquad$ .
(1) 0 Hint: From $m+\frac{1}{m}=1$, we get $m^{2}-m+1=0$, thus
$$
m^{3}+1=(m+1)\left(m^{2}-m+1\right)=0,
$$

which means $m^{3}=-1$. Therefore,
$$
m^{2008}+\frac{1}{m^{2009}}=\left(m^{3}\right)^{669} \cdot m+\frac{m}{\left(m^{3}\right)^{670}}=-m+m=0 .
$$