Question

Question #e6daa

Answer 1

The complex conjugate of `Re^(itheta)` is `Re^(-itheta)`

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

Proceeding,

`(Re^(itheta))^"*"color(white)(_)=(R(cos(theta)+isin(theta)))^"*"`

`=(Rcos(theta)+iRsin(theta))^"*"`

`=Rcos(theta)-iRsin(theta)`

`=R(cos(theta)-isin(theta))`

`=R(cos(-theta)+isin(-theta))`

`=Re^(i(-theta))`

`=Re^(-itheta)`