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Question Number 9255 by geovane10math last updated on 26/Nov/16

e^{i π} + 1 = 0

e^{i π} = - 1

{(e^{i π})}^{2} = {(- 1)}^{2}

e^{2 i π} = 1

{(e^{2 i π})}^{i} = 1^{i}

1^{i} = e^{2 i^{2} π}

1^{i} = e^{- 2 π}

1^{i} = 1^{a + b i} ? ? ?

x^{a + b i} = ? ? ?

Commented by RasheedSoomro last updated on 26/Nov/16

“ Squaring both {sides}^{″} may not yield equivalent

statement .

Commented by FilupSmith last updated on 27/Nov/16

e^{2 i π} = 1

e^{2 i π} = \cos (2 π) + i \sin (2 π)

= 1 + i (0) = 1