两边取模得|z+i|^m=|z-i|^m,当m>0时|z+i|=|z-i|,设z=x+yi,x,y∈R,则x^+(y+1)^=x^+(y-1)^,化简得4y=0,y=0,∴z∈R.若m为正整数,由二项式定理,∑c(m,k)z^(m-k)[i^k-(-i)^k]=0,∴∑c(m,2p+1)z^(m-2p-1)*2i^(2p+1)=0....