1 = 1 -1=1 ?

Algebra Level 4

I. Start with the famous identity e i π = 1 e^{iπ}=-1 .

II. So, e π = ( 1 ) 1 / i e^π=(-1)^{1/i} .

III. So, e π / i = ( 1 ) ( 1 / i ) × ( 1 / i ) = ( 1 ) 1 = 1 e^{π/i}=(-1)^{(1/i) \times (1/i)}=(-1)^{-1}=-1 .

IV. So, e π / i = e i π e^{π/i}=e^{iπ} .

V. So, π / i = i π π/i=iπ or i 2 = 1 i^2=1 .

VI. But i = 1 i=\sqrt{-1} or i 2 = 1 i^2=-1 . So, 1 = 1 -1=1 .

In which step is there an error?

V II IV III VI I

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2 solutions

Antonio Hugo
Mar 5, 2016

Since e^iθ is periodic function, we aren't allowed to compare the exponent. It's similar with trig function. sinθ=sinα,θ≠α

Yes, you are correct.

A Former Brilliant Member - 5 years, 3 months ago

However the very first error is in step III. because the brackets around -1 are lost!!!

Andreas Wendler - 5 years, 3 months ago

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I thought the same thing…

Dallin Richards - 5 years, 3 months ago

how can we transfer the i in step 2???

Darvin Rio - 5 years, 2 months ago
Ruofeng Liu
Dec 6, 2019

e a = e b a = b e^a=e^b \iff a=b only holds true for real numbers

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