Sum of Complex Numbers in Exponential Form/General Result

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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:

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


Proof

Let:

\(\displaystyle r e^{i \theta}\) \(=\) \(\displaystyle \displaystyle \sum_{k \mathop = 1}^n z_k\)
\(\displaystyle \) \(=\) \(\displaystyle z_1 + z_2 + \dotsb + z_k\)
\(\displaystyle \) \(=\) \(\displaystyle r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2} + \dotsb + r_n \paren {\cos \theta_n + i \sin \theta_n}\) Definition of Complex Number
\(\displaystyle \) \(=\) \(\displaystyle r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}\) rerranging

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

\(\displaystyle r\) \(=\) \(\displaystyle \sqrt {\map {\Re^2} z + \map {\Im^2} z}\)
\(\displaystyle r\) \(=\) \(\displaystyle \sqrt {\paren {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n }^2 + \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}^2 }\)

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


\((1):\quad\) \(\, \displaystyle 1 \le k \le n: \, \) \(\displaystyle {r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2\) \(=\) \(\displaystyle {r_k}^2 \paren {\cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle {r_k}^2\) Sum of Squares of Sine and Cosine
\((2):\quad\) \(\, \displaystyle 1 \le j < k \le n: \, \) \(\displaystyle 2 r_j r_k \cos \theta_j \cos \theta_k + 2 {r_j} {r_k} \sin \theta_j \sin \theta_k\) \(=\) \(\displaystyle 2 r_j r_k \paren {\cos \theta_j \cos \theta_k + \sin \theta_j \sin \theta_k}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 r_j r_k \, \map \cos {\theta_j - \theta_k}\) Cosine of Difference


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:

\(\displaystyle \map \tan \theta\) \(=\) \(\displaystyle \frac y x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\map \Im z} {\map \Re z}\)

Hence:

\(\displaystyle \theta\) \(=\) \(\displaystyle \map \arctan {\frac {\map \Im {r e^{i \theta} } } {\map \Re {r e^{i \theta} } } }\)
\(\displaystyle \) \(=\) \(\displaystyle \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }\)

$\blacksquare$


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