# Sum of Complex Numbers in Exponential Form

## Theorem

Let $z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$ be complex numbers expressed in exponential form.

Let $z_3 = r_3 e^{i \theta_3} = z_1 + z_2$.

Then:

$r_3 = \sqrt { {r_1}^2 + {r_2}^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$
$\theta_3 = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2} {r_1 \cos \theta_1 + r_2 \cos \theta_2} }$

### General Result

Let $n \in \Z_{>0}$ be a positive integer.

For all $k \in \set {1, 2, \dotsc, n}$, let:

$z_k = r_k e^{i \theta_k}$

Let:

$r e^{i \theta} = \ds \sum_{k \mathop = 1}^n z_k = z_1 + z_2 + \dotsb + z_k$

Then:

 $\ds r$ $=$ $\ds \sqrt {\sum_{k \mathop = 1}^n {r_k}^2 + \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \map \cos {\theta_j - \theta_k} }$ $\ds \theta$ $=$ $\ds \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }$

## Proof

We have:

 $\ds r_1 e^{i \theta_1} + r_2 e^{i \theta_2}$ $=$ $\ds r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2}$ Definition of Polar Form of Complex Number $\ds$ $=$ $\ds \paren {r_1 \cos \theta_1 + r_2 \cos \theta_2} + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2}$

Then:

 $\ds {r_3}^2$ $=$ $\ds {r_1}^2 + {r_2}^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2}$ Complex Modulus of Sum of Complex Numbers $\ds \leadsto \ \$ $\ds r_3$ $=$ $\ds \sqrt { {r_1}^2 + {r_2}^2 + 2 r_1 r_2 \map \cos {\theta_1 - \theta_2} }$

and similarly:

$\theta_3 = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2} {r_1 \cos \theta_1 + r_2 \cos \theta_2} }$

$\blacksquare$