Question #e6daa

1 Answer
Aug 27, 2016

Answer:

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)#