# Simplifying $\exp {- i 2 \pi / N}$.

a and b are complex numbers and I know the equation below.

$$X_{N} = a + e^{-i2\pi /N}*b$$

I wanted to simplify it. Here is what I've tried.

I know $e^{-i\pi} = -1$

$X_{N} = a + \left ( e^{i\pi} \right )^{-2/N}*b$

$X_{N} = a + \left ( -1 \right )^{-2/N}*b$

$X_{N} = a + \left ( 1 \right )^{-1/N}*b$

$X_{N} = a + b$

$X_{N} = a + \frac{1}{\sqrt[N]{1}}*b$

I know this wrong but I do not know why? Do you have any idea?

Another question: How can I correctly simplify it?

The complex power functions is defined by $x^y=e^{y\ln x}$ where $\ln$ is the $complex$ logarithm. For this it does not hold in general that $(e^a)^b=e^{ab}$ for $a,b\in \mathbb{C}$. The problem is that the complex logarithm is multivalued (it's the inverse of the exponential which is not injective), so you must agree on some principal value.

So not all properties of the real exponential function are carried over to the complex case. Another well-known example where this goes wrong is

$1=\sqrt{1}=\sqrt{-1\cdot-1}=(\sqrt{-1})^2=i^2=-1$.

Or you can use Euler's identity: $e^{ix}=\cos x+i\sin x$
Then $a+be^{-i\frac{2\pi}{N}}=a+b \ \cos\left(\frac{2\pi}{N}\right)-i b\sin\left(\frac{2\pi}{N}\right)$