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

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 7:22
• 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 7:23
• 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 7:27

3 Answers

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

• So only if $b,c$ are integers, then $(a^b)^c = a^{bc}$ always? – Joshhh Apr 12 '16 at 7:26
• @Joshhh Not "only if". If $a, b, c$ are all positive, for instance, then that rule holds. – Patrick Stevens Apr 12 '16 at 8:01

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.

• But $1,\theta,2\pi$ are all real numbers. – Joshhh Apr 12 '16 at 7:29
• They are not necessarily integers, except $1$. In fact, $2\pi$ is not an integer clearly. – choco_addicted Apr 12 '16 at 7:30
• $1^x$ 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 7:32
• 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 7:36
• ${i\theta} = {i\frac{2\pi}{2\pi}\theta} = \log_e (e^{i2\pi})^{\frac{\theta}{2\pi}} = \log_e 1^{\frac{\theta}{2\pi}} = 0$ ... at log level – Narasimham Apr 12 '16 at 7:37

$\theta$ 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 $2\pi$