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

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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