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

be non-zero complex numbers in exponential form.

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

Then:

Proof
Let:

By the definition of the complex modulus, with $z = x + i y$, $r$ is defined as:
 * $r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}$

Hence

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

Hence:


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

Note that $r > 0$ since $r_k > 0$ for all $k$.

Hence we may safely assume that $r > 0$ when determining the argument below.

By definition of the argument of a 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 $z$.

As $r > 0$ we have that $\cmod z \ne 0$ by definition of modulus.

Hence we can divide $(2)$ by $(1)$, to get:

Hence: