$-1=1$ ?

Algebra Level 4

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

II. So, $e^π=(-1)^{1/i}$ .

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

IV. So, $e^{π/i}=e^{iπ}$ .

V. So, $π/i=iπ$ or $i^2=1$ .

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

In which step is there an error?

V II IV III VI I

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

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

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