Is there a way of adding two vectors in polar form without first having to convert them to cartesian or complex form?

Here is another way forward that relies on straightforward vector algebra. Let $\vec r_1$ and $\vec r_2$ denote vectors with magnitudes $r_1$ and $r_2$, respectively, and with angles $\phi_1$ and $\phi_2$, respectively.

Let $\vec r$ be the vector with magnitude $r$ and angle $\phi$ that denotes the sum of $\vec r_1$ and $\vec r_2$. Thus,

From the definition of the inner product we have

and

Using $(1)$ and $(2)$, we find $r^2$ can be written

and thus $r$ is given by

Using $(1)$, $(3)$, and $(4)$, yields

whereupon solving for $\cos (\phi-\phi_1)$ reveals

We can easily obtain the expression for $\sin(\phi-\phi_1)$ by applying the cross product

which after straightforward arithmetic yields

Dividing $(5)$ by $(6)$ and inverting shows that

where the function $\operatorname{arctan2}(y,x)$ is described in this article.

Equations $(4)$ and $(7)$ provide the polar coordinates of $\vec r$ strictly in terms of the polar coordinates of $\vec r_1$ and $\vec r_2$. And the development of $(4)$, $(5)$, and $(6)$ did not appeal to Cartesian coordinates.

NOTE:

In a parallel development, we can express the sum of two complex numbers $z_1=r_1e^{i\phi_1}$ and $z_2=r_1e^{i\phi_2}$ in terms of their magnitudes and arguments.

First, recall that the inner product of two complex numbers is given by

where $\bar z$ denotes the complex conjugate of $z$.

Next, we let $z=re^{i\phi}=z_1+z_2$ be the sum of $z_1$ and $z_2$. The magnitude of $z$ is given by

Therefore, we have

Finally, we find the argument of $z$ by taking the inner product of $z$ and $z_1$. To that end, we write

which reveals that

whereupon inverting yields

Equations $(8)$ and $(9)$ provide the polar coordinates of $z$ strictly in terms of the polar coordinates of $z_1$ and $z_2$. Again, this development did not appeal to Cartesian coordinates.