Electronic – How does one measure the power factor

So, in general case, the electric field should be kind of: $$E_3=A\sqrt{\frac{1+b\cos(2\omega_h t)}{2}}\text{e}^{j\omega_c t}$$ You may worry about the sign '\$\pm\$'before the square root. That would be a very good consideration and we need to extend our solution into a generic expression. We may notice the amplitude factor could be extended as: $$\pm\sqrt{\frac{1+b\cos(2\omega_h t)}{2}}\\ =\pm\sqrt{\frac{1+b[2\cos^2(\omega_h t)-1]}{2}}\\ =\cos(\omega_h t)\cdot\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}} $$ It is easy to verify that for \$b=1\$ the above expression is just \$\cos(\omega_ht)\$ as we desired. So we can rewrite \$E_3\$ as: $$ E_3=\frac{A\cdot\beta(t)}{2}\cdot\bigg[\text{e}^{j(\omega_c-\omega_h)t}+\text{e}^{j(\omega_c+\omega_h)t}\bigg]\;, $$ where \$\beta(t)=\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}}\$. Maybe...Maybe... You also worry about the singularities of \$\beta(t)\$ $$\cos(\omega_ht)\rightarrow0\quad\Rightarrow\quad\beta(t)\rightarrow\infty\;.$$ To cure this problem, I introduce a prescriptive parameter \$\epsilon\$, it could be an arbitrary small positive number (like 0.01,0.001) in the denominator for avoiding the divergence. $$ \beta(t)\rightarrow\beta'(t)=\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)+\epsilon}} $$ My next step is to find the Fourier spectrum of '\$\cos(\omega_h t)\sqrt{b+\frac{1-b}{2\cos^2(\omega_h t)}}\$' between \$ -\omega_h\$ to \$\omega_h\$--because we already have the phase '\$\text{e}^{j\omega_c t}\$'in our expression. This will be useful for getting monochrome modes.

Here are the pictures I made

for b=1

for b=0.5

$$-------------------------------------$$ Here is the code for Matlab

Th=20; %% period of omega_h
omegah=2*pi/Th;%%% we define \omegah via defining Th
N=90; %% divide the time interval (one period) into N equal parts.
L=100; %% Length of time interval interms of numbers of periods.
tt=Th/N*[1:L*N]; %% we generate a time vector including 3 periods.
b=0.5;
amph=cos(omegah*tt).*sqrt(b+(1-b)*power((2*cos(omegah*tt).*cos(omegah*tt)),-1)); %% this is our factor in the amplitude.
plot(tt,amph) %%% the picture of the amplitude including 'L' periods.
ttrial=Th/N*[1:3*N];
amphtrial=cos(omegah*ttrial).*sqrt(b+(1-b)*power((2*cos(omegah*ttrial).*cos(omegah*ttrial)),-1));
plot(ttrial,amphtrial) %%% the picture of the amplitude including '3' periods, this picture just helps you tell the pattern of the mode easily.
%%
FThh=fft(amph);
Homg=fftshift(FThh); %%% if you were not familiar of usage of commands 'fft' and 'fftshift' check them in the matlab help, they are extremely useful!!!
angleomg=-pi*N/Th:2*pi/(L*Th):pi*N/Th-2*pi/(L*Th);
plot(angleomg,abs(Homg))
title('b=0.5')
xlabel('\omega')
ylabel('E')

A definite mistake is your method of calculating total RMS voltage. In the absence of anything else other than "240V" and "21V", you have to assume these are RMS but you have assumed they are peak.

You could argue that you are right but not really, because you used them as RMS voltages earlier on to calculate currents and these will also be RMS values that you don't need to divide by sqrt(2) either.

Other than that and me not understanding the term "displacement power factor" and "true power factor" you are correct.