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我不知道怎麼翻譯Jhone Tsan Chang
Jan 12, 2021

請問 0^i=? e^(i ln0) ln0=-∞? e^(-∞i)=??? 還有ln(-1)=?,有沒有複數解?

in Math Problems
1 answer0 replies
1
joe li
Jan 15, 2021

i^i=e^ln(i^i)

=e^ilni

=e^iln(sqrt(-1))

=e^(i/2)ln(-1)

we also know by euler's formula

i^i=(e^((pi/2)i))^i because i=e^((pi/2)i)

i^i=e^-pi/2

so

e^-pi/2=e^(i/2)ln(-1)

-pi/2=iln(-1)/2

ln(-1)=(pi)i


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