Theorem: Sum of Complex Numbers in Exponential Form/General Result

Let $n \in \Z_{>0}$ be a positive integer.
For all $k \in \set {1, 2, \dotsc, n}$, let:
:$z_k = r_k e^{i \theta_k}$
be non-zero complex numbers in exponential form.
Let:
:$r e^{i \theta} = \ds \sum_{k \mathop = 1}^n z_k = z_1 + z_2 + \dotsb + z_k$
Then:
{{begin-eqn}}
{{eqn | l = r
      | r = \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} }
}}
{{eqn | l = \theta
      | r = \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} }
}}
{{end-eqn}}

Proof:
Let:
{{begin-eqn}}
{{eqn | l = r e^{i \theta}
      | r = \sum_{k \mathop = 1}^n z_k
      | c =
}}
{{eqn | r = z_1 + z_2 + \dotsb + z_k
      | c =
}}
{{eqn | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2} + \dotsb + r_n \paren {\cos \theta_n + i \sin \theta_n}
      | c = {{Defof|Complex Number}}
}}
{{eqn | r = r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}
      | c = rerranging
}}
{{end-eqn}}
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
{{begin-eqn}}
{{eqn | l = r
      | r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}
      | c =
}}
{{eqn | l = r
      | r = \sqrt {\paren {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n }^2 + \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}^2 }
      | c =
}}
{{end-eqn}}
In the above we have two types of pairs of terms:
{{begin-eqn}}
{{eqn | n = 1
      | q = 1 \le k \le n
      | l = {r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2
      | r = {r_k}^2 \paren {\cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2}
      | c =
}}
{{eqn | r = {r_k}^2
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | n = 2
      | q = 1 \le j < k \le n
      | l = 2 r_j r_k \cos \theta_j \cos \theta_k + 2 {r_j} {r_k} \sin \theta_j \sin \theta_k
      | r = 2 r_j r_k \paren {\cos \theta_j \cos \theta_k + \sin \theta_j \sin \theta_k}
      | c =
}}
{{eqn | r = 2 r_j r_k \map \cos {\theta_j - \theta_k}
      | c = Cosine of Difference
}}
{{end-eqn}}
Hence:
:$\ds r = \sqrt {\sum_{k \mathop = 1}^n {r_k}^2 + \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \map \cos {\theta_j - \theta_k} }$
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:
{{begin-eqn}}
{{eqn | l = \map \tan \theta
      | r = \frac y x
      | c =
}}
{{eqn | r = \frac {\map \Im z} {\map \Re z}
      | c =
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \theta
      | r = \map \arctan {\frac {\map \Im {r e^{i \theta} } } {\map \Re {r e^{i \theta} } } }
      | c =
}}
{{eqn | r = \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} }
      | c =
}}
{{end-eqn}}
{{qed}}