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Write down w* in polar form

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi guys I have been given a question, write down w* in polar form where w=2< -(pi/3). I can work out the question when it is in cartesian form just not this way, any help woud be great.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 18, 2009 #2

    arildno

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    What does "<" mean??
     
  4. Oct 18, 2009 #3
    Not entirely sure thats just the way it is shown in the question. All i know is when i converted it to cartesian form it became 1 - SQRT3 i
     
  5. Oct 18, 2009 #4

    HallsofIvy

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    You have left two things ambiguous. As arilno implied "<" is not a standard notation but I am going to assume that you meant the complex number is written in the polar form with modulus r= 3 and angle, or "argument", [itex]\theta= \pi/3[/itex].

    The other thing that is ambiguous is the *. I am going to assume that you mean "complex conjugate" which is more commonly written [itex]\overline{w}[/itex].

    The connection between "Cartesian representation" and "polar representation" is [itex]z= x+ iy= r (cos(\theta)+ i sin(\theta))[/itex] or, equivalently, [itex]z=x+ iy= r e^{i\theta}[/itex] The complex conjugate is gotten, basically, by changing the sign on "i":
    [itex]\overline{z}= x- iy= r (cos(\theta)- i sin(\theta))[/itex] which, because cosine is an even function and sine is an odd function, can be written [itex]\overline{z}= x- iy= r (cos(\theta)- i sin(\theta))[/itex][itex]= cos(-\theta)+ i sin(\theta)[/itex].

    Similarly, from [itex]z= x+ iy= r e^{i\theta}[/itex], [itex]\overline{z}= x- iy= r e^{-i\theta}[/itex].

    In either case, the complex number given by modulus r and argument [itex]\theta[/itex] has complex conjugate given by modulus r and argument [itex]-\theta[/itex].
     
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