- #1
T@P
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ok don't kill me, hear me out before you say its undefined:
[tex] e ^ { \pi i}= -1 [/tex]
so taking the natural log of both sides yields:
[tex] ln(-1) = \pi i [/tex]
[tex] \frac {ln(-1)} \pi = i [/tex]
and plugging it into the well known equation,
[tex] e^{\varphi i} = cos(\varphi) + i sin(\varphi) [/tex]
you substitute your i into the left side,
[tex] e^{\frac {ln (-1)} { \pi * \varphi}} = cos(\varphi) + i sin(\varphi)[/tex]
[tex] -1^{\frac {\varphi} { \pi}} = cos(\varphi) + i sin(\varphi)[/tex]
now the left side is arguably real, so if you solve for i, you have i in terms of a real number. some of the time. i realize that you phi/pi can be 1/2 which would get you back to i, but can anyone explain this to me?
thanks (excuse the shoddy latex)
ok for some reason instead of getting something to the power of i get a funny ) sign. please assume it means to the power of, like this sign: ^
[tex] e ^ { \pi i}= -1 [/tex]
so taking the natural log of both sides yields:
[tex] ln(-1) = \pi i [/tex]
[tex] \frac {ln(-1)} \pi = i [/tex]
and plugging it into the well known equation,
[tex] e^{\varphi i} = cos(\varphi) + i sin(\varphi) [/tex]
you substitute your i into the left side,
[tex] e^{\frac {ln (-1)} { \pi * \varphi}} = cos(\varphi) + i sin(\varphi)[/tex]
[tex] -1^{\frac {\varphi} { \pi}} = cos(\varphi) + i sin(\varphi)[/tex]
now the left side is arguably real, so if you solve for i, you have i in terms of a real number. some of the time. i realize that you phi/pi can be 1/2 which would get you back to i, but can anyone explain this to me?
thanks (excuse the shoddy latex)
ok for some reason instead of getting something to the power of i get a funny ) sign. please assume it means to the power of, like this sign: ^
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