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We have: $$ \Re\{Ae^{i\phi}e^{i\omega}\}=\Re\{Ae^{i(\phi+\omega)}\}=A\cos(\phi+\omega) $$ But why does the following not give the same the answer: \begin{align} \Re\{Ae^{i\phi}e^{i\omega}\} &=A\Re\{e^{i\phi}e^{i\omega}\}\\ &=A\Re\{e^{i\phi}\}\Re\{e^{i\omega}\}\\ &=A\cos(\phi)\cos(\omega) \quad ? \end{align}

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    $\begingroup$ $\Re(ab)\ne\Re(a)\Re(b)$. $\endgroup$ – Cave Johnson Dec 28 '17 at 13:05
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You made the assumption in the penultimate line that the real part of a product is a product of the real parts. But that's simply not true. Remember the definition for complex multiplication in rectangular coordinates?

Say let us have $w=a+ib$ and $z=p+iq$, then $$Re(wz)=ap-bq\neq{ap}=Re(w)Re(z)$$ as you assumed.

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As @CaveJohnson wrote in a comment, the real part of a complex number is not a multiplicative function. An easy counter-example would be

$$\Re(i)\Re(i)=0\neq-1=\Re(i^2)$$

Hence, only your first calculation is correct.

EDIT: You can calculate directly and see that in fact, if $z_{1}=a+bi$ and $z_{2}=c+di$, then

$$\Re\left(z_{1}z_{2}\right)=\Re\Big((ac-bd)+(ad+bc)i\Big)=ac-bd=\Re(z_{1})\Re(z_{2})-\color{red}{\Im(z_{1})\Im(z_{2})}$$

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  • $\begingroup$ Hi! I checked en.wikipedia.org/wiki/Complex_number, but I couldn't find any rule why it is wrong? $\endgroup$ – JDoeDoe Dec 28 '17 at 13:10
  • $\begingroup$ @JDoeDoe Why do you assume it's true? Wikipedia doesn't list this as a statement. $\endgroup$ – eranreches Dec 28 '17 at 13:12
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    $\begingroup$ @JDoeDoe You're into a wrong way of thinking. In mathematics, only what can be proved is considered true; what has not yet been disproved is only a conjecture. Let alone the fact that it has been disproved in this answer. For an easy example, can you find any source claiming $192\times 364=69886$ is wrong? Does this mean that it is a correct equality? $\endgroup$ – Cave Johnson Dec 28 '17 at 13:18

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