Which is the error? $e^{\frac{2\pi i}{3}}=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}$ o $e^{\frac{2\pi i}{3}}=1$.

Question

Which is the error?

We know that $$e^{\frac{2\pi i}{3}}=-\frac{1}{2}+ i\frac{\sqrt{3}}{2}.$$

What is the error in the following argument?

$$e^{\frac{2\pi i}{3}}=\left(e^{\pi i}\right)^{\frac{2}{3}}=(-1)^{\frac{2}{3}}=\left((-1)^{2}\right)^{\frac{1}{3}}=(1)^{\frac{1}{3}}=1.$$

So with the link supplied in the previous comment, your error is in the first equality. — Laars Helenius, Apr 13 '17 at 03:42
@Laars but the first equality is correct, but by chance (it is not true in all cases but it is true in this specific case). — , Apr 13 '17 at 03:44
I find it interesting that Matlab gives the same result for $e^{i2\pi/3}$ and $(e^{i\pi})^{2/3}$. — Cye Waldman, Apr 13 '17 at 19:26

score 1 · Answer 1 · 2017-04-13T03:49:40.783

1

The error is that in general

$$a^{p/q}\neq (a^p)^{1/q}$$

The reason is that $a^{p/q}$ is defined as $(a^{1/q})^p$.

edited Apr 13 '17 at 03:49

answered Apr 13 '17 at 03:43

1 Answers1