√i ??????

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)

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