# Sum of 2 complex exponentials

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1. Oct 5, 2015

### wirefree

I appreciate the opportunity afforded by this forum to submit a question.

I have struggled with the derivation shown in the attached picture. I am certainly unfamiliar with the concept used to include the arctan function in the encircled step.

Would be highly appreciative of a prompt.

Best regards,
wirefree

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2. Oct 5, 2015

### BvU

Hello there,

I then assume you are happy with (4) and wonder how the factor in [ ] brackets is converted to the polar form in the red oval.
For simplicity I propose to introduce $$v = A\cos\alpha+B\cos\beta \\ w = A\sin\alpha+B\sin\beta$$ so that we have to convert $v + jw$ to polar form, i.e. to a form $R\,e^{j\phi}$

(which is the reverse of what was done to write out (4), where, for example, they wrote$$A\,e^{j\alpha}= A\cos\alpha + j A\sin\alpha$$)

If we want to solve $v + jw = R\,e^{j\phi}$ for $R$ and $\phi$ and compare with the above, we see that $\phi$ follows from ${\sin\phi\over\cos \phi} = {w\over v}$, meaning $\tan\phi = w/v$ or $\phi = \arctan {w\over v}$

R follows from $R^2 = (v+jw)(v-jw) = v^2 - j^2 w^2 = v^2 + w^2$

What helps a lot in this is to draw a picture of the complex plane (j axis vertical, real axis horizontal). And a unit circle or a circle with radius R to convert $R \, e^{j\phi}$ to real and imaginary parts.

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3. Nov 14, 2015

### wirefree

Thank you, BvU.