k EF=∑/(kz=∑(2z2 K=0 K=0 1-(2z 1-2Z11-2z1 2z|<1,Z|>2F(Z)= ︱2Z ︱＜1 , ︱Z︱＞2 ( ) ( ) k K K  f k  Z  =  = − = 0 0 -k 1 Z 2 ( ) 1 1 1 1 2 1 1 2 1 2 − − −  − = − − = Z Z Z