Pole locations of Butterworth filter
Assuming that you understand the left-hand side of Eq. $(8.47)$, for understanding the right-hand side you need to know that $-1=e^{j\pi}$, and that $e^{j2k\pi}=1$. So in order to obtain all $2N$ roots of $(-1)^{\frac{1}{2N}}$ you rewrite $-1$ as
$$-1=e^{j\pi}e^{j2\pi k}\tag{1}$$
from which you get
$$(-1)^{\frac{1}{2N}}=e^{j\frac{\pi}{2N}}e^{j\frac{2\pi k}{2N}}=e^{j\frac{\pi}{2N}(1+2k)},\qquad k=0,1,\ldots,2N-1\tag{2}$$
And since $j=e^{j\pi/2}$, the final result is
$$\Omega_cj(-1)^{\frac{1}{2N}}=\Omega_ce^{j\pi/2}e^{j\frac{\pi}{2N}(1+2k)}=\Omega_ce^{j\frac{\pi}{2N}(1+2k+N)},\qquad k=0,1,\ldots,2N-1\tag{3}$$
If this is all new to you, you should really read up on complex numbers. They are fundamental in understanding many concepts in signal processing.

PS: There is a typo in the equations in the book. The numerator on the right-hand of Eq. $(8.46)$ should have $\Omega_c$ instead of $\Omega$. Also, the left-hand side of Eq. $(8.47)$ should have $\Omega_c$ instead of $\Omega$.