Sum of Complex Numbers in Exponential Form/General Result

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

Then:

Proof
Let:

From Definition:Complex Modulus, with $z = x + i y$, $r$ is defined as $\sqrt {{\map {\Re^2} z} + {\map {\Im^2} z} }$

Hence

In the above we have two types of pairs of terms:


 * $(1) \, 1 \le k \le n$
 * ${r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2 = {r_k}^2 \paren { \cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2 } = {r_k}^2$ Sum of Squares of Sine and Cosine


 * $(2) \, 1 \le j \lt k \le n$
 * $2 {r_j} {r_k} \cos {\theta_j} \cos {\theta_k} + 2 {r_j} {r_k} \sin {\theta_j} \sin {\theta_k} = 2 {r_j} {r_k} \paren { \cos {\theta_j} \cos {\theta_k} + \sin {\theta_j} \sin {\theta_k} } = 2 {r_j} {r_k} \cos \paren { {\theta_j} - {\theta_k} }$ Cosine of Difference

Hence


 * $r = \sqrt { \displaystyle \sum_{k \mathop = 1}^n r_k + \displaystyle \sum_{1 \le j \lt k \le n} 2 {r_j} {r_k} \cos \paren { {\theta_j} - {\theta_k} } }$

Note:


 * $r = 0$ only when all $r_k = 0$. In the case where $r = 0$ then the argument of the complex number isn't defined, so we may safely assume that $r \gt 0$ when determining the argument below.

From Definition:Argument of Complex Number, with $z = x + i y$, $\theta$ is defined as any solution to the pair of equations:
 * $(1): \quad \dfrac x {\cmod z} = \map \cos \theta$
 * $(2): \quad \dfrac y {\cmod z} = \map \sin \theta$

where $\cmod z$ is the modulus of $z$.

Dividing $(2)$ by $(1)$, when $\cmod z \ne 0$ (see Note above) gives:

Hence