Question #e6daa

Aug 27, 2016

The complex conjugate of $R {e}^{i \theta}$ is $R {e}^{- i \theta}$

Explanation:

Standard notation for the complex number $\sqrt{- 1}$ is $i$, and so that will be used in the answer.

Given a complex number $w$ with polar form $R {e}^{i \theta}$, we wish to find the complex conjugate ${w}^{\text{*}}$ of $w$ in polar form. To do so, we will use Euler's formula : ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, along with the facts that the sine function is odd and the cosine function is even.

Proceeding,

${\left(R {e}^{i \theta}\right)}^{\text{*"color(white)(_)=(R(cos(theta)+isin(theta)))^"*}}$

$= {\left(R \cos \left(\theta\right) + i R \sin \left(\theta\right)\right)}^{\text{*}}$

$= R \cos \left(\theta\right) - i R \sin \left(\theta\right)$

$= R \left(\cos \left(\theta\right) - i \sin \left(\theta\right)\right)$

$= R \left(\cos \left(- \theta\right) + i \sin \left(- \theta\right)\right)$

$= R {e}^{i \left(- \theta\right)}$

$= R {e}^{- i \theta}$