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Transfer Function Magnitude, how to deal with 'j'?

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I have the simple transfer function of an RC filter:

\$ H(s) = \frac{sRC}{1 + sRC} \$

In order to find the magnitude, square the previous equation and take the square root:

\$ (H(s))^2 = \frac{(sRC)^2}{(sRC + 1)^2} ; s = j\omega \$

\$ (H(s))^2 = \frac{(j\omega R C)^2}{(j\omega R C + 1)^2} \qquad(1) \$

This is where I get stuck. I know that the correct magnitude should be of the form:

\$ |H(s)| = \sqrt{\frac{(\omega R C)^2}{(\omega R C)^2 + 1}} \qquad (2) \$

However, when I actually try to calculate this out on paper I don't know how to deal with the \$j\$. I don't understand how to get from equation I get from equation (1) to equation (2). The \$-1\$ from \$j^2\$ causes the form to look like the following:

\$ (H(s))^2 = \frac{-top}{1 - bottom} \$

If I were to take the square root of this I do not get equation (2).

What am I doing wrong?


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