Help with $C\cos(\omega t+\phi )=\Re{\Big\{De^{i\omega t}\Big\}}$

Question

I have a question regarding the following form \begin{align} x(t)&=A\cos(\omega t)+B\sin (\omega t)\tag 1\\ &=C\cos(\omega t+\phi )\tag 2\\ &=\Re{\Big\{De^{i\omega t}\Big\}}\tag 3\\ &=\Re{\Big\{Ee^{i(\omega t+\phi)} \Big\}} \tag 4 \end{align}

For a complex number $z=a+ib$ I know I can write \begin{align} z=a+ib=\sqrt{a^2+b^2}e^{i\phi} =\lvert z\rvert e^{i\phi} \tag 5 \end{align}

Question:

I guess I have $D=Ce^{i\phi}$? Should I compare this with equation $(5)$, so I actully have $$ D=a+ib=\sqrt{a^2+b^2}e^{i\phi}=Ce^{i\phi}=\lvert D\rvert e^{i\phi} \tag 6 $$ So $C=\sqrt{a^2+b^2}=\lvert D\rvert$? Is this correct?

Answer 1

Yes. It is correct. Let $cos(\phi) = A/\sqrt{A^2+B^2}$ and $sin(\phi) = -B/\sqrt{A^2+B^2}$

then $x(t) =\sqrt{A^2+B^2}[cos(\omega t)cos(\phi) - sin(\omega t)sin(\phi)]$

$ = \sqrt{A^2+B^2}cos(\omega t + \phi)$

$= \Re{\Big\{\sqrt{A^2+B^2}e^{i(\omega t + \phi)} \Big\}} $. Let $C=\sqrt{A^2+B^2}$.

$= \Re{\Big\{Ce^{i(\omega t + \phi)} \Big\}} = \Re{\Big\{Ce^{i\phi}e^{i(\omega t )} \Big\}} = \Re{\Big\{De^{i(\omega t )} \Big\}} $

So $D= Ce^{i\phi}$, and

$C = |C|= |D| = \sqrt{A^2+B^2} $.