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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.

Adding two polar vectors

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.

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    $\begingroup$ Hint: to compute $z_1 - z_2$, you can compute $z_1 + (-z_2)$. So if you can negate $z_2$ (in polar form), then you can apply the addition formula from the other page to get the result you want. Hint: to negate $(r, \theta)$, try $(r, \theta + \pi)$. $\endgroup$ Commented Jul 23, 2018 at 16:26
  • $\begingroup$ Thanks for the hint and the quick answer, I will try your approach. $\endgroup$
    – Pavel
    Commented Jul 25, 2018 at 19:12

1 Answer 1

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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.

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    $\begingroup$ Thank you for the quick answer. I get the point with the transformation. Maybe I can rearrange the expression when putting the $\phi_2$ and $\phi_1$ in relation. For example when both phase values are quite close to each other. Thanks again $\endgroup$
    – Pavel
    Commented Jul 25, 2018 at 19:10

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