Does $1^{\frac{-i\ln 2}{2\pi}}$ equal 2? Just out of curiosity, I would like to know if this derivation is correct or not.
Let's assume complex numbers and write $1 = e^{2\pi i n}$, for any $n\in\mathbb{Z}$. 
Then, by exponentiation we obtain $$1^{\frac{-i\ln 2}{2\pi}}=e^{2\pi i n \cdot \frac{-i \ln 2}{2\pi}} = 2^n,$$ and thus for $n=1$, $1=2$.
For me, this looks like a big contradiction. Any power of $1$ should be equal to $1$, or? What is the catch here that I don't see? In complex numbers, the power of $1$ doesn't have to be equal to $1$? Thanks.
 A: The rule $(a^b)^c=a^{bc}$ does not hold for complex numbers.
A: For $k \in \mathbb{Z}$ and $z \in \mathbb{C}$, 
$$1^z = (e^{i2\pi k})^z = e^{i2\pi k \cdot z}$$
If $z = x$ where $x \in \mathbb{R}$, then $1^z = e^{i2\pi k \cdot x}$ can be any number on on the complex unit circle. Note that the $x$ is rotating the point $z=1$ around the circle.
If $z = x + iy $ where $x,y \in \mathbb{R}$, then $1^z = e^{-y}e^{i2\pi k \cdot x}$ can be any number on the complex plane except zero. Again we have a rotation by $x$. But now the magnitude can assume any non-vanishing value, depending on $y$.
In a nutshell, $1^z$ does not always equal 1 when dealing with complex numbers. Spooky! You may be interested in trying to find what sort of $z$ gives $1^z = 1$. 
A: $$1^{\frac{-i\ln 2}{2\pi}}=$$
$$\left|1^{\frac{-i\ln 2}{2\pi}}\right|e^{\arg\left(1^{\frac{-i\ln 2}{2\pi}}\right)i}=$$
$$|1|^{\frac{-i\ln 2}{2\pi}}e^{\arg\left(1^{\frac{\ln(2)}{2\pi}e^{-\frac{1}{2}\pi i}}\right)i}=$$
$$\sqrt{1^2}e^{\arg\left(1\right)i}=$$
$$\sqrt{1}e^{0i}=$$
$$1e^{0i}=$$
$$e^{0i}=$$
$$1(\cos(0)+\sin(0)i)=\cos(0)+\sin(0)i=1+0i=1$$
