Let $\theta \in (-\pi,\pi) $ we have that $$ e^{i\theta} = \left(e^{i\theta}\right)^\frac{2\pi}{2\pi} = (e^{2\pi i})^{\frac{\theta}{2\pi}} $$ so as $e^{2\pi i}=1$ we have that $$ e^{i\theta} = (1)^{\frac{\theta}{2\pi}} =1 $$ But there must be something wrong because $e^{i\pi}=-1 \neq 1$.
Asked
Active
Viewed 33 times
0
-
Cf. this question – J. W. Tanner Mar 21 '24 at 23:19
-
2Yes, so by having the $\frac{1}{2}$ power, you're actually taking a square root, right? Well, the square root of 1 is normally 1 by convention, but really there are two solutions to this, yes? -1 or 1. Truly, the issue here seems to be that you said squaring a number and taking the square root always gives you that value back, but in truth it gives you the absolute value of that number back, i.e. turning -1 into 1 – cable Mar 21 '24 at 23:20