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

**Answer**

Here is another way forward that relies on straightforward vector algebra. Let →r1 and →r2 denote vectors with magnitudes r1 and r2, respectively, and with angles ϕ1 and ϕ2, respectively.

Let →r be the vector with magnitude r and angle ϕ that denotes the sum of →r1 and →r2. Thus,

→r=→r1+→r2

From the definition of the inner product we have

→r1⋅→r2=r1r2cos(ϕ2−ϕ1)

and

→r1⋅→r=r1rcos(ϕ−ϕ1)

Using (1) and (2), we find r2 can be written

r2=→r⋅→r=(→r1+→r2)⋅(→r1+→r2)=→r1⋅→r1+→r2⋅→r2+2→r1⋅→r2r21+r22+2r1r2cos(ϕ2−ϕ1)

and thus r is given by

\bbox[5px,border:2px solid #C0A000]{r=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}}\tag 4

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

\begin{align}

\vec r_1\cdot \vec r&=r_1r\cos(\phi-\phi_1)\\\\

&=\vec r_1\cdot (\vec r_1+\vec r_2)\\\\

&=r_1^2+r_1r_2\cos(\phi_1-\phi_2)\\\\

\end{align}

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

\cos(\phi-\phi_1)=\frac{r_1+r_2\cos(\phi_2-\phi_1)}{\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}}\tag 5

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

\hat z\cdot(\vec r_1 \times \vec r)=\hat z\cdot(\vec r_1 \times \vec r_2)

which after straightforward arithmetic yields

\sin(\phi-\phi_1)=\frac{r_2\sin(\phi_2-\phi_1)}{\sqrt{r_1^2+r_2^2+2r_1r_2\cos(\phi_2-\phi_1)}} \tag 6

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

\bbox[5px,border:2px solid #C0A000]{\phi =\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right)} \tag 7

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

\begin{align}

\langle z_1,z_2 \rangle &=z_1 \bar z_2\\\\

&=r_1r_2e^{i(\phi_1-\phi_2)}

\end{align}

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

\begin{align}

r&=\sqrt{\langle z,z \rangle}\\\\

&=\sqrt{\langle z_1+z_2,z_1+z_2 \rangle}\\\\

&=\sqrt{r_1^2+r_2^2+r_1r_2\left(e^{i(\phi_1-\phi_2)}+e^{-i(\phi_1-\phi_2)}\right)}\\\\

&=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}

\end{align}

Therefore, we have

\bbox[5px,border:2px solid #C0A000]{r=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}} \tag 8

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

\begin{align}

\langle z,z_1 \rangle &=rr_1e^{i(\phi-\phi_1)}\\\\

&=\langle z_1+z_2,z_1 \rangle \\\\

&=r_1^2+r_1 r_2 e^{i(\phi_2-\phi_1)}

\end{align}

which reveals that

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 9

whereupon inverting yields

\bbox[5px,border:2px solid #C0A000]{\phi =\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right)}

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.

**Attribution***Source : Link , Question Author : lash , Answer Author : Mark Viola*