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#487 [–¼‘O‚È‚µ]
e^{iz}=(ã2}1)i
e^{iz}=(ã2}1){cos{(1/2+2n)ƒÎ}+isin{(1/2+2n)ƒÎ}}
=e^{log(ã2}1)}e^{(1/2+2n)ƒÎi}
=e^{log(ã2}1)+(1/2+2n)ƒÎi}
iz=log(ã2}1)+(1/2+2n)ƒÎi
z=-ilog(ã2}1)+(1/2+2n)ƒÎ
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:09/02/11 12:30 :PC :™™™
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