√i ??????
回答 (2)
2006-12-20 11:28 pm
✔ 最佳答案
利用隸美弗定理 (De Moivre's Theorem) 和複數極式 (Polar Form of complex number).先將 i 寫成:
i = cos(pi/2) + isin (pi/2)
Then
√i = √[cos(pi/2) + isin (pi/2)]
= [cos(pi/2) + isin (pi/2)]^(1/2)
= cos(pi/4) + isin (pi/4) (De Moivre's Theorem)
= (1/√2) + (i/√2)
= (1/√2) (1+i)
參考: My Maths knowledge
2006-12-20 11:37 pm
let (a+bi)^2=i
(a+bi)^2=a^2+2abi-b^2
therefore
i=a^2+2abi-b^2
By comparing like terms,
a^2+b^2=0 ...(1)
2ab=1 .........(2)
From (1),
we know
b=a ............(3)
Sub (3) into (2)
we have
2a^2=1
a=b=2^ -0.5
therefore (2^ -0.5+2^ -0.5i)^2 =√i
(a+bi)^2=a^2+2abi-b^2
therefore
i=a^2+2abi-b^2
By comparing like terms,
a^2+b^2=0 ...(1)
2ab=1 .........(2)
From (1),
we know
b=a ............(3)
Sub (3) into (2)
we have
2a^2=1
a=b=2^ -0.5
therefore (2^ -0.5+2^ -0.5i)^2 =√i
收錄日期: 2021-04-12 22:40:49
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