Why does moving the imaginary (\$i\$/\$j\$) component in the capacitor element of a transfer function to the numerator flip the coefficient's sign?
This is more of a question for math.stackexchange, but remember that \$j = \sqrt{-1}\$. Therefore, \$\frac{1}{j} = \frac{1}{\sqrt{-1}}= \frac{\sqrt{-1}}{-1}=-\sqrt{-1}=-j\$. Note that I multiplied by \$\frac{\sqrt{-1}}{\sqrt{-1}}=1\$ to get rid of the square root in the denominator.