Question #e6daa

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Question #e6daa

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Answer 1 · 2016-08-27T07:51:40

The complex conjugate of $R e^{i θ}$ $Re^(itheta)$ is $R e^{- i θ}$ $Re^(-itheta)$

Explanation:

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

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

Proceeding,

${(R e^{i θ})}^{*}_= {(R (cos (θ) + i sin (θ)))}^{*}$ $(Re^(itheta))^"*"color(white)(_)=(R(cos(theta)+isin(theta)))^"*"$

$= {(R cos (θ) + i R sin (θ))}^{*}$ $=(Rcos(theta)+iRsin(theta))^"*"$

$= R cos (θ) - i R sin (θ)$ $=Rcos(theta)-iRsin(theta)$

$= R (cos (θ) - i sin (θ))$ $=R(cos(theta)-isin(theta))$

$= R (cos (- θ) + i sin (- θ))$ $=R(cos(-theta)+isin(-theta))$

$= R e^{i (- θ)}$ $=Re^(i(-theta))$

$= R e^{- i θ}$ $=Re^(-itheta)$

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Question #e6daa

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