isn't 1^i=1? and using that logic, isn't i to the i not e^-pi/2?

by Rory Cannon   Last Updated January 16, 2018 09:20 AM

isn't 1^i=1? because i^i = (i x 1)^i = i^i x 1^i = i ^i? or we could use e^PIi instead because Ill be disproving i to the i later e^PIi = -1 1^PI = 1 1^PIi = 1^i e^PIi x 1^i = e^PIi x 1^PIi e^PIi x 1^PIi = (e x 1)^PIi (e x 1)^PIi = e^PIi = -1 e^PIi x 1^i = e^PIi 1^i = 1

that is simple algebra

i^i = x i^4i = x^4 i^4i = (i^4)^i x^4 = 1^i x^4 = 1 x = 1, -1, i, or -i

simple algebra

e^PIi = -1 e^2PIi = 1 e^-2PI = 1^i e^-2PI = 1

somebody disprove this

simple algebra



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