問題 1.4 Laurent展開

問題 1.4 は巻末の略解が誤っている部分なので注意が必要です.

問題は以下の通り(p.11*1 )

問題 1.4 \mathbb{C}_z上の有理型函数\phi(z)=\frac{z}{(z+1)(z^2-1)}の極と極の点でのローラン展開を求めよ.

極とは分母がゼロになるところと考えればよいので, 極はz= \pm 1になりますね.

\phi(z)=\frac{z}{(z+1)^2(z-1)}とすこしお色直しをしておきましょう.

(i)z=-1の近くでの様子

\phi(z)
=\frac{z}{(z+1)^2(z-1)}
=\frac{\{(z+1)-1\}} {(z+1)^2\{(z+1)-2\} }
=\frac{\zeta-1}{\zeta^2(\zeta-2)} (\zeta=z+1とした)
=\zeta^{-2}(\zeta-1)\frac{1}{\zeta-2}
=\zeta^{-2}(\zeta-1)(-\frac{1}{2})\frac{1}{1-(\frac{1}{2}\zeta)}
!=!\zeta^{-2}(\zeta-1)(-\frac{1}{2})\sum_{n=0}^{\infty}(\frac{1}{2}\zeta)^{n} (ここの等号は局所的です)
=(\zeta-1)(-\frac{1}{2})\sum_{n=0}^{\infty} (\frac{1}{2})^{n}\zeta^{n-2}
=(-\frac{1}{2})\{\sum_{n=0}^{\infty} (\frac{1}{2})^{n}\zeta^{n-1}-\sum_{n=0}^{\infty} (\frac{1}{2})^{n}\zeta^{n-2}\}
=(-\frac{1}{2})\{\sum_{n=-1}^{\infty} (\frac{1}{2})^{n+1}\zeta^{n}-\sum_{n=-2}^{\infty} (\frac{1}{2})^{n+2}\zeta^{n}\}
=(-\frac{1}{2})\{-\zeta^{-2}+(1-1/2)\zeta^{-1}+\sum_{n=0}^{\infty} ((\frac{1}{2})^{n+1}-(\frac{1}{2})^{n+2})\zeta^{n}\}
=(\frac{1}{2})\zeta^{-2}+(-\frac{1}{4})\zeta^{-1}+(-\frac{1}{2})\sum_{n=0}^{\infty} ((\frac{1}{2})^{n+1}-(\frac{1}{2})^{n+2})\zeta^{n} =(\frac{1}{2})\zeta^{-2}+(-\frac{1}{4})\zeta^{-1}+(-\frac{1}{2})\sum_{n=0}^{\infty} (\frac{1}{2})^{n+2}(2-1)\zeta^{n}
=(\frac{1}{2})\zeta^{-2}+(-\frac{1}{4})\zeta^{-1}+\sum_{n=0}^{\infty} -(\frac{1}{2})^{n+3}\zeta^{n}
=(\frac{1}{2})(z+1)^{-2}+(-\frac{1}{4})(z+1)^{-1}+\sum_{n=0}^{\infty} -(\frac{1}{2})^{n+3}(z+1)^{n}

計算確認を www.wolframalpha.com を使ってしてみました.

(i)z=1の近くでの様子

\phi(z)
=\frac{z}{(z+1)^2(z-1)}
=\frac{\{(z-1)+1\}} {\{(z-1)+2\}^2(z+1)}
=\frac{\zeta+1}{(\zeta+2)^2\zeta} (\zeta=z-1とした)
=(\zeta)^{-1}(\zeta+1)(\frac{1}{\zeta+2})^2
=(\zeta)^{-1}(\zeta+1)(-\frac{1}{\zeta+2})^{\prime}
=(\zeta)^{-1}(\zeta+1)(-\frac{1}{2}\frac{1}{1-(\frac{1}{2}\zeta)})^{\prime}
=-\frac{1}{2}(\zeta)^{-1}(\zeta+1)(\frac{1}{1-(\frac{1}{2}\zeta)})^{\prime}
!=!-\frac{1}{2}(\zeta)^{-1}(\zeta+1)(\sum_{n=0}^{\infty}(-\frac{1}{2}\zeta)^{n})^{\prime} (ここの等号は局所的です) =-\frac{1}{2}(\zeta)^{-1}(\zeta+1)(\sum_{n=1}^{\infty}-\frac{1}{2}n(-\frac{1}{2}\zeta)^{n-1})
=-\frac{1}{2}(\zeta)^{-1}(\zeta+1)(\sum_{n=1}^{\infty}-\frac{1}{2}n(-\frac{1}{2})^{n-1}\zeta^{n-1})
=(\zeta+1)(\sum_{n=1}^{\infty}n(-\frac{1}{2})^{n+1}\zeta^{n-2})
=\sum_{n=1}^{\infty}n(-\frac{1}{2})^{n+1}\zeta^{n-1}+\sum_{n=1}^{\infty}n(-\frac{1}{2})^{n+1}\zeta^{n-2} =\sum_{n=0}^{\infty}(n+1)(-\frac{1}{2})^{n+2}(\zeta)^{n}+\sum_{n=-1}^{\infty}(n+2)(-\frac{1}{2})^{n+3}(\zeta)^{n} =\sum_{n=0}^{\infty}(n+1)(-\frac{1}{2})^{n+2}(\zeta)^{n}+1(-\frac{1}{2})^{(-1)+3}\zeta^{-1}+\sum_{n=0}^{\infty}(n+2)(-\frac{1}{2})^{n+3}(\zeta)^{n} =\frac{1}{4}\zeta^{-1}+\sum_{n=0}^{\infty}\{(n+1)(-\frac{1}{2})^{n+2}+(n+2)(-\frac{1}{2})^{n+3}\}\zeta^{n} =\frac{1}{4}\zeta^{-1}+\sum_{n=0}^{\infty}(-\frac{1}{2})^{n+3}\{-2(n+1)+(n+2)\}\zeta^{n} =\frac{1}{4}\zeta^{-1}+\sum_{n=0}^{\infty}-(-\frac{1}{2})^{n+3}(-n)\zeta^{n}
=\frac{1}{4}\zeta^{-1}+\sum_{n=0}^{\infty}n(-\frac{1}{2})^{n+3}\zeta^{n}
=\frac{1}{4}(z-1)^{-1}+\sum_{n=0}^{\infty}n(-\frac{1}{2})^{n+3}(z-1)^{n}

計算確認を www.wolframalpha.com を使ってしてみました.

なかなか骨の折れる計算でした.

*1:小木曽啓示　(2002)　「代数曲線論」　朝倉書店