Sum of Complex Numbers in Exponential Form

From ProofWiki
Jump to: navigation, search

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:

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


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:

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


Then:

\(\displaystyle r\) \(=\) \(\displaystyle \sqrt { {r_1}^2 + {r_2}^2 + \dotsb + {r_n}^2 + \ldots}\) $\quad$ $\quad$
\(\displaystyle \theta\) \(=\) \(\displaystyle \map \arctan {\dfrac {r_1 \sin \theta_1 + r_1 \sin \theta_2 + \dotsb + r_n \sin \theta_2} {r_1 \cos \theta_1 + r_1 \cos \theta_2 + \dotsb + r_n \cos \theta_2} }\) $\quad$ $\quad$


Proof

We have:

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

Then:

\(\displaystyle {r_3}^2\) \(=\) \(\displaystyle r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2}\) $\quad$ Complex Modulus of Sum of Complex Numbers $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle r_3\) \(=\) \(\displaystyle \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2} }\) $\quad$ $\quad$


and similarly:

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

$\blacksquare$


Sources