$e^{i\theta} = 1$ for all $\theta$?

Question

What is the conceptual error in the following derivation? $$ e^{i\theta} = e^{i\frac{2\pi}{2\pi}\theta} = (e^{i2\pi})^{\frac{\theta}{2\pi}} = 1^{\frac{\theta}{2\pi}} = 1 $$ It is clear to me that the second move is illegal, I just don't know why.

I suspect that $(a^b)^c\neq a^{bc}$ in general in $\mathbb{C}$. — Justin Benfield, Apr 12 '16 at 07:22
I bet it's something like (-8)^(1/3) = -2, (-8)^(1/3) != (-8)^(2/6) = 64^(1/6) = 2 — S.Klumpers, Apr 12 '16 at 07:23
I've got another head scratcher for you :
$$1=\sqrt{1}=\sqrt{(-1)^2}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i\cdot i = i^2=-1.$$ — Mathematician 42, Apr 12 '16 at 07:27

score 2 · Answer 1 · answered Apr 12 '16 at 07:24

2

$a^{x y} \not = (a^x)^y$ in general.

For example, $[(-1)^2]^{1/2} = 1$ but $(-1)^1 = -1$.

You may wish to look up "branch cuts" to understand conceptually why this is true.

answered Apr 12 '16 at 07:24

Patrick Stevens

36,135

So only if $b,c$ are integers, then $(a^b)^c = a^{bc}$ always? – Joshhh Apr 12 '16 at 07:26
1

@Joshhh Not "only if". If $a, b, c$ are all positive, for instance, then that rule holds. – Patrick Stevens Apr 12 '16 at 08:01

choco_addicted · Answer 2 · 2016-04-12T07:55:47.580

1

First, the formula $$ e^{i\frac{2\pi}{2\pi}\theta}=(e^{i2\pi})^{\theta/2\pi} $$ is not correct, because $z^{\alpha\beta}=(z^{\alpha})^{\beta}$ does not hold, in general.

Second, the formula $$ 1^{\theta/2\pi}=1 $$ is not correct.

Complex power is defined as $$ a^b = e^{b\ln a} $$ and can have (in)finitely many values if $b$ is not a real integer.

If $z^{\alpha\beta}=(z^{\alpha})^{\beta}$ holds, then $\beta$ is an integer. For detailed information, see here, page 12.

edited Apr 12 '16 at 07:55

answered Apr 12 '16 at 07:28

choco_addicted

8,786
8
24
53

But $1,\theta,2\pi$ are all real numbers. – Joshhh Apr 12 '16 at 07:29
They are not necessarily integers, except $1$. In fact, $2\pi$ is not an integer clearly. – choco_addicted Apr 12 '16 at 07:30
$1^x$ is a constant function, so $1^{\frac{\theta}{2\pi}}$ is necessarily $1$. Why is the second move I made not true in general? – Joshhh Apr 12 '16 at 07:32
In real, your statement is true. But in complex analysis it is different. Compute $1^{1/2}$:$$1^{1/2}=e^{\frac{1}{2}\ln 1} =e^{\frac{1}{2} 2k\pi i} = e^{k\pi i}=(-1)^k,$$ where $k$ is an integer. – choco_addicted Apr 12 '16 at 07:36
$ {i\theta} = {i\frac{2\pi}{2\pi}\theta} = \log_e (e^{i2\pi})^{\frac{\theta}{2\pi}} = \log_e 1^{\frac{\theta}{2\pi}} = 0 $ ... at log level – Narasimham Apr 12 '16 at 07:37

anni saini · Answer 3 · 2016-04-20T10:19:31.347

-1

$\theta$ is the angle from the real axis so what you did that while converting $e^{i\theta}$ to $e^{i\theta \frac{2\pi}{2\pi}}$ you change the complex number because if $\theta \ne 2\pi$ then $e^{i\theta \frac{2\pi}{2\pi}} \ne e^{i2\pi \left(\frac{\theta}{2\pi}\right)}$ because the argument of the $e^{i\theta }$ is $\theta$ and the argument of $e^{i2\pi}$ is $2\pi$

edited Apr 20 '16 at 10:19

answered Apr 12 '16 at 07:38

anni saini

259

DeMoivre's formula works only for integer powers. – Alex M. Apr 17 '16 at 10:01

$e^{i\theta} = 1$ for all $\theta$?

3 Answers3