# Polar Form Addition without Rectangular Form?

As I've learned, and also described this answer that

In rectangular form, complex numbers are easy to add; just add their components.

In polar form, complex numbers are easy to multiply; just multiply their magnitudes and add their arguments.

Thou the polar form presents an easier way to multiply two values, would it even be possible to add two polar form values without transforming them to the rectangular form? If so, how?

The answers in this question (How to add real number and complex number in polar form) are still somewhat particular. I am looking of a generic addition in the phasor form, such as,

$$(A_1 \angle \theta_1 ) + (A_2 \angle \theta_2 ) = ?$$

• Let's forget for a moment about the without rectangular form requirement: We get a (not very nice-looking) formula. Now, there's not going to be a different formula if we forget about rectangular form. Does that observation not answer the question? Commented Feb 25, 2022 at 18:50
• You can transform to rectangular form, add, transform back. You get a formula for generic addition in the phasor form that you can now use without ever converting to rectangular form but the formula is not very pleasant. Commented Feb 25, 2022 at 18:52
• On my application, the issue is that going to and back from the rectangular form would include an angle calculation (e.g. arctan functions). I am wondering if there would be a not-very-nice-looking formulas could contain some type of trigonometric tricks that would avoid such operations. Commented Feb 25, 2022 at 18:55
• As Biil Cook's answer shows, no there is no such formula, i.e., trig does not work in one's favor here.... Commented Feb 25, 2022 at 21:23

You can extract the rectangular parts from the polar expression, add, and then re-express to do this, but it ain't pretty.

Recall that if $$z=a+bi$$ where $$a,b \in \mathbb{R}$$, then the modulus is $$|z|=\sqrt{a^2+b^2}$$ and the argument is $$\mathrm{arg}(z)$$ set of angles associated with $$z$$'s polar form.

$$z=|z|e^{\mathrm{arg}(z)i}=|z|\cos(\mathrm{arg}(z))+i|z|\sin(\mathrm{arg}(z))$$

$$w+z=(|z|\cos(\mathrm{arg}(z))+|w|\cos(\mathrm{arg}(w)))+i(|z|\sin(\mathrm{arg}(z))+|w|\sin(\mathrm{arg}(w)))$$

Then the modulus of $$w+z$$ is $$\sqrt{(|z|\cos(\mathrm{arg}(z))+|w|\cos(\mathrm{arg}(w)))^2+(|z|\sin(\mathrm{arg}(z))+|w|\sin(\mathrm{arg}(w)))^2}$$.

The argument is trickier to extract. Say a representative of the argument lies strictly between $$-\pi/2$$ and $$\pi/2$$ (in particular, the imaginary part isn't zero), you could use arctangent to get at a representative angle: $$\mathrm{arctan}\left(\frac{|z|\sin(\mathrm{arg}(z))+|w|\sin(\mathrm{arg}(w))}{|z|\cos(\mathrm{arg}(z))+|w|\cos(\mathrm{arg}(w))}\right)$$.

In the end, it's nasty.

• Thank you for confirming that out. I was really hoping for something different than arctan. Commented Feb 25, 2022 at 22:23