Find $z=1+\epsilon+\cdots+\epsilon^{n-1}$ where $\epsilon^{2n}=1$, $n\in \mathbb{N},\epsilon\in \mathbb{C}$

Question

Find $z=1+\epsilon+\cdots+\epsilon^{n-1}$ where $\epsilon^{2n}=1$, $n\in \mathbb{N},\epsilon\in \mathbb{C}$

Solution to the equation $\epsilon^{2n}=1$ is $\epsilon=1$.

Could someone explain how?

Then, $z$ should be $z=n-1$.

Is this correct?

Answer 1

As you already noted, $\epsilon=1$ is a solution to $\epsilon^{2n}=1$. With that value, you get $z=n$.

If $\epsilon\ne 1$, you can rewrite your equation as $$z = \frac{\epsilon^n-1}{\epsilon-1}$$ Now we have $(\epsilon^n)^2 = \epsilon^{2n} = 1$, therefore we can distinguish two cases here: Either $\epsilon^n=1$ of $\epsilon^n=-1$.

In the case $\epsilon^n=1$, obviously $z=0$ (because the numerator is zero, but the denominator isn't, since we are still considering the case $\epsilon\ne 1$).

The interesting case is when $\epsilon^n=-1$. In that case, we get $$z = \frac{2}{1-\epsilon}$$

Now the most general solution of $\epsilon^{2n}=1$ is $\epsilon = \mathrm e^{\mathrm i\pi k/n}$ with $k\in \mathbb Z$ (note that actually we can restrict consideration to $0\le k<2n$ because of the $2\pi\mathrm i$ periodicity of the exponential function). Since we are considering the case that $\epsilon^n=-1$, $k$ must be odd, that is, $k=2m+1$ for some $m\in \mathbb Z$. So the most general solution in that case is $$z = \frac{2}{1 - \mathrm e^{\mathrm i\pi (2m+1)/n}}$$ It is a good idea to expand the fraction with $\mathrm e^{-\mathrm i\pi (2m+1)/(2n)}$, so that in the denominator we get twice a sine. Then we get $$z = -\frac{\mathrm e^{-\mathrm i\pi(2m+1)/(2n)}}{\sin\left(\pi\frac{2m+1}{2n}\right)} = -\frac{\cos\left(\pi\frac{2m+1}{2n}\right)}{\sin\left(\pi\frac{2m+1}{2n}\right)}+\mathrm i = \mathrm i-\cot\left(\pi\frac{2m+1}{2n}\right)$$

Answer 2

If $\varepsilon=1$ then $1+\varepsilon+\varepsilon^2+\cdots+\varepsilon^{n-1} = n$, not $n-1$.

Notice that $\varepsilon=-1$ is another solution to $\varepsilon^{2n}=1$, and in that case $1+\varepsilon+\varepsilon^2+\cdots+\varepsilon^{n-1} = 0$ if $n$ is even (since then you have equally many $1$s and $-1$s). If $n$ is odd, then $n-1$ is even, and so the number of $1$s is $1$ more than the number of $-1$s and the sum is $1$.

However, it is probable that whoever wrote this question had in mind that $\varepsilon$ is a not-necessarily-real complex number. If $\varepsilon\in\{ \pm1, \pm i\}$, then $\varepsilon^{2n}=1$ when $n=2$. In that case, the sum $1+\varepsilon+\varepsilon^2+\cdots+\varepsilon^{n-1}$ has only one term and the sum is $1$.

If $n=3$, then $\dfrac{1+i\sqrt 3} 2 \vphantom{\dfrac \int {\displaystyle \int}}$ and all of its powers are solutions of $\varepsilon^{2n}=1$. In that case we have $1+\varepsilon+\varepsilon^2$ $= 1+i\sqrt 3 = 2\varepsilon$.

In general, $\varepsilon = \cos\dfrac\pi n + i\sin\dfrac \pi n = e^{i\pi/n}$ is a solution to $\varepsilon^{2n}=1$. We have \begin{align} & 1+\varepsilon+\varepsilon^2+\cdots+\varepsilon^{n-1} = \frac{\varepsilon^n-1}{\varepsilon-1} \qquad (\text{sum of a geometric series}) \\[10pt] = {} & \frac{-2}{\varepsilon - 1} = \frac 2 {1 - e^{i\pi/n}} \\[10pt] = {} & \frac{2(1-e^{-i\pi/n})}{(1-e^{i\pi/n})(1-e^{-i\pi/n})} \qquad(\text{The conjugage of }e^{i\pi/n}\text{ is } e^{-i\pi/n}.) \\[10pt] = {} & \frac{2(1-e^{-i\pi/n})}{2-2\cos(\pi/n)} = e^{-i\pi/(2n)} \frac{e^{i\pi/(2n)} - e^{-i\pi/(2n)}}{1-\cos(\pi/n)} = e^{-i\pi/(2n)} \frac{2i\sin(\pi/(2n))}{1-\cos(\pi/n)}. \end{align}

Other solutions to $\varepsilon^{2n}=1$ are powers of the one above; they are $$ \left( \cos\frac\pi n + i \sin \frac \pi n \right)^k = \cos \frac{k\pi} n + i \sin \frac{k\pi} n. $$ What happens with exponents $k$ other than $1$ might bear examination.