ㄧ題簡單留數問題~~~

sinhz=z+z^3/3!+z^5/5!+...
1/(1-z^2)=1+z^2+z^4+z^6+...
so, sinhz/(1-z^2)=(z+z^3/6+...)(1+z^2+z^4+...)
=z+(7/6)z^3+...(higher order than z^4)
so, residue of sinhz/[z^4(1-z^2)] at pole z=0 is 7/6

恩...
因為解別人的發問者會看有錯可以切磋
可是自己解只有自己看

版大會解http://tw.knowledge.yahoo.com/question/question?qid=1509122805656
為何還問這題?