My textbook wants to solve for poles of the following polynomial
$$ \frac{1}{1+(s/j\Omega_c)^{2N}} $$
A pole will be where
\begin{align*} 1+(s/j\Omega_c)^{2N} &= 0 \\ s &= (-1)^{1/2N} j \Omega_c \\ \end{align*}
From there my book says there will be $2N$ poles of the form
\begin{align*} s_k &= \Omega_c e^{(j\pi/2N)(2k+N-1)} \\ \end{align*}
for $k = 0, 1, \ldots, 2N-1$.
How do I derive this step from the previous step. If I substitute $-1=e^{j\pi}$ and $j=e^{j\pi/2}$, I get:
\begin{align*} s &= \Omega_c \cdot e^{j\pi/2N} e^{j\pi/2} \\ s &= \Omega_c \cdot e^{(j\pi/2N) (1 + N)} \\ \end{align*}
How do I get from that to the book derived answer?
Let us start from the elementary stuff: $$ -1 = cos(m\pi)+ i \sin(m\pi)$$ where $m$ can be any odd integer i.e. $m = 2k-1$ where $k \in \mathcal{Z}$. To find, say, $p$-th root of it we write: $$ (-1)^{1/p} = \left(cos((2k-1)\pi)+ i \sin((2k-1)\pi)\right)^{1/p} = cos((2k-1)\pi/p)+ i \sin((2k-1)\pi/p) = e^{j(2k-1)\pi/p}.$$ To get $p$ different roots we consider $k = 0,1,2,3,\ldots, (p-1)$. It is easy to see that roots will start repeating when we put $k = p$ or $k = p+1$ and so on. In this problem we have to replace $p$ by $2N$ to get all the possible roots of $-1$ as required in the problem. Thus we have $$ (-1)^{1/2N}= e^{j(2k-1)\pi/2N},$$ where $k = 0,1,2,3,\ldots, (2N-1)$, Hence the $s$-value ($s = (-1)^{1/2N} j \Omega_c$), corresponding to $k$-th root will be: \begin{align*} s_k & = e^{j(2k-1)\pi/2N} j \Omega_c \\ & = e^{j(2k-1)\pi/2N} e^{j\pi/2} \Omega_c \\ & = e^{j(2k-1)\pi/2N + jN\pi/2N} \Omega_c \\ & = e^{(j\pi/2N) (2k+N-1)} \Omega_c \end{align*} where $k = 0,1,2,3,\ldots, (2N-1)$.