CVS4

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Find the Laurent series about z_{0}=1\, for the function {\frac  {e^{{2z}}}{(z-1)^{3}}}\,

The only singularity is at z=1\, so the series will look like

\sum _{{n=-\infty }}^{{\infty }}A_{n}(z-1)^{n},0<|z-1|<\infty \,

A_{n}={\frac  {1}{2\pi i}}\oint _{C}{\frac  {f(z)}{(z-1)^{{n+1}}}}dz={\frac  {1}{2\pi i}}\oint _{C}{\frac  {e^{{2z}}}{(z-1)^{{n+4}}}}dz,n=0,\pm 1,\pm 2,...\,

where C\, is any simple closed curve around the point z=1\,.

Cauchy's theorem gives A_{n}=0,n=-4,-5,...\,

For n\geq 3\,

A_{n}={\frac  {1}{(n+3)!}}{\frac  {d^{{n+3}}}{dz^{{n+3}}}}e^{{2z}}{\Bigg |}_{{z=1}}={\frac  {e^{2}2^{{n+3}}}{(n+3)!}},n\geq 3\,.

The Laurent series is

f(z)={\frac  {e^{{2z}}}{(z-1)^{3}}}=8e^{2}\sum _{{n=-3}}^{\infty }{\frac  {2^{n}}{(n+3)!}}(z-1)^{n},0<|z-1|<\infty \,.


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