ãã解ï¼æçæ°çå®ä¹æ¥æ±è§£ãå¨z=0å¤ççæ°ï¼å³f(z)å¨z=0å±å¼æ´æ级æ°ä¸ææ°ä¸º-1项çç³»æ°c-1ã
ããâµsin(1/z)=1/z-1/(3!z^3)+1/(5!z^5)+â¦â¦ï¼(z-1)^2=z^2-2z+1ï¼
ããâ´f(z)=ï¼z-1)^2sin(1/z)=(z^2-2z+1)[1/z-1/(3!z^3)+1/(5!z^5)+â¦â¦ï¼]=z-2+(1-1/6)/z+1/(3z^2)+â¦â¦ï¼â´c-1=5/6ï¼å³Res[f(z),0]=5/6ãä¾åèã
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