Real part of $\exp(i\theta + i\phi)$

Question

I need to determine the real part of

$$e^{i(\theta+\phi)}$$

I first used

$$Re(e^{i(\theta+\phi)})=\cos(\theta+\phi)$$

However, $e^{i(\theta+\phi)}$ can also be written as $e^{i\theta}*e^{i\phi}$. Now, when I take the real part of this expression, I get

$$Re(e^{i\theta}*e^{i\phi})=\cos(\theta)\cos(\phi)$$

Now, using the trigonometric rule : $ \cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

We see that $$Re(e^{i(\theta+\phi)})\neq Re(e^{i\theta}*e^{i\phi})$$

However, this shouldn't be the case, since $e^{i(\theta+\phi)}$ and $e^{i\theta}*e^{i\phi}$ are equal and should therefore have the same real part.

Is there a mistake I made, or something I am not taking into account?

Thanks in advance for your time!

kind regards, Suzanne

Answer 1

The issue here is that $\Re (e^{i\theta} e^{i\phi}) = \cos \theta \cos \phi$ is false; you have $e^{i\theta} = \cos \theta + i \sin \theta$ and $e^{i\phi} = \cos \phi + i \sin \phi$, and multiplying these gives $\cos \theta \cos \phi - \sin \theta \sin \phi + i(\cos \theta \sin \phi + \sin \theta \cos \phi)$, which has real part $\cos \theta \cos \phi - \sin \theta \sin \phi$ (which happens to be $\cos (\theta + \phi)$).

Answer 2

You did miss something. The real part is actually

$$\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)$$

as pure imaginary * pure imaginary = real