[理工] [工數] 複變
題目: find all sulution of sin z = i
e^iz - e^-iz
解:1. 因 sin z = i 故 -------------- = i
2i
則 e^2zi + 2 e^iz -1 = 0
︿︿︿︿︿︿︿︿︿︿︿
1
答案: z = 2nπ + ----ln(-1 + √2 ) n=( 0 . +-1 . +-2 ......)
i -
這怎嚜出來的?
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※ 編輯: bizzard 來自: 114.35.149.203 (02/01 08:59)
※ 編輯: bizzard 來自: 114.35.149.203 (02/01 09:26)
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