Understanding unexpected i/q calculation on a downsampled signal
Because the decimate function does not try to keep power constant.

Remember that decimation is not inherently an LTI operation; so, probably, the authors of that function simply did not care too much for keeping the amplitude of a passband signal constant.

Note that a decimation step typically includes filtering to $\frac1{\text{ds_factor}}$ of the original bandwidth to avoid aliases. That filter might or might not have a passband gain $\le 1$.

A general remark:


  For each pair in (for i,q in zip(x_real, x_imag)), I calculate h=math.sqrt(i**2 + q**2). I don't expect h to ever be larger than sqrt(2), because the real and imaginary parts are normalized to [-1, 1].


In fact, if coming right out of a receiver with good IQ balance, don't expect $h\ge1$, even!

Remember that I and Q aren't independent; they are the inphase and quadrature component of a signal whose max power, no matter which phase, is limited! Thus, as long as you don't risk clipping in your receiver, $i^2 \le {1-q^2}\iff q^2\le 1-i^2$! 

If your I and Q would be independently limited to an amplitude of 1, your $\sqrt 2$ would be right. They are not – imagine, for example, the passband signal that gives you $i\equiv1$: it's the cosine with the carrier frequency $1\cdot\cos(2\pi f_c t + 0)$ (note the $0$ phase). Now, any cosine with larger power would clip, right? Obviously, cosines of the same amplitude would also be at the extreme power limit, no matter whether they have zero phase. The moment the phase is not zero, you'll see some of the cosine's energy appear on Q, and missing on I. However, it's the same power.