digital IQ modulators when it comes to digital I/Q signal power measurement we know that this power is viewed as $I^2+Q^2$ (as its polar form). This is the power as referred to final analogue signal. Yet the IQ modulator does not convert I/Q to polar form, instead only the final I is output to real world say in the RF modulator.
When I model two I/Q mixers it does not show that final power is $I^2 + Q^2$ but rather just $I^2$:

x = randn(1,1024)+j*randn(1,1024); %random I/Q input
sincos1 = exp(j*2*pi*(0:1023)*.01); 
y1 = x.*sincos1; % first digital IQ modulator 

sincos2 = exp(j*2*pi*(0:1023)*.1);
y2 = real(y1.*sincos2); %rf IQ mixer giving final analogue output


if you check power of y2 you will see it is close to real(y1) rather than complex y1.
I ask this because for all power dependent applications e.g. DPD we assume power is $I^2+Q^2$
what am I missing here?

Thanks
The power of the final real-valued signal $y_2$ is half the power of $y_1$ because you take the real part (and thus reduce power by a factor 2). The real part of $y_1$ is already the sum of the I-component and the Q-component, not only the I-component:

$$Re\{y_1\}=Re\{x\}\cos\omega_1 t - Im\{x\}\sin\omega_1 t= I-Q$$

So the original formula $I^2+Q^2$ is correct, you just need to apply a factor of $1/2$ if you don't compensate for taking the real-part. This is why in some textbooks the real part operation in a bandpass modulator is 'implemented' as $\sqrt{2}Re\{\cdot\}$; this will keep the power unchanged.