I tried to subtract two complex numbers in polar form without transforming them into the cartesian form. Therefore I used the approach made by Mark Viola in the following link.
I managed to get the following result.
$$e^{i(\phi-\phi_1)}=\frac{r_1-r_2e^{i(\phi_2-\phi_1)}}{\sqrt{r_1^2+r_2^2-2r_1r_2\cos (\phi_2-\phi_1)}} \tag 1$$
At this point I do not know how to achieve the final equation like mentioned in the link. I post the final equation for adding two complex numbers in polar form:
$${\phi=\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right)} \tag 2$$
Any help or hints which leading to the final equation are appreciated.
Best Answer
In fact, you can't avoid the conversion from polar to Cartesian and back to polar, even if done in a single go (any expression you will find will be of a comparable complexity).
The combined equations are
$$r^2=(r_1\cos\phi_1-r_2\cos\phi_2)^2+(r_1\sin\phi_1-r_2\sin\phi_2)^2 \\=r_1^2-2r_1r_2\cos(\phi_1-\phi_2)+r_2^2$$
and
$$\tan\phi=\frac{r_1\sin\phi_1-r_2\sin\phi_2}{r_1\cos\phi_1-r_2\cos\phi_2}.$$
I don't think there is a nice simplification of the expression of the argument. And computationally, the simplification of the modulus is not so attractive as it involves an additional evaluation of a circular function.