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Simplifying and then expressing complex numbers in cartesian form

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    (2 CIS (pi/6))*(3 CIS (pi/12))

    2. Relevant equations

    Also what is CIS? I believe it's Cos+i*sin but how do you use it?


    3. The attempt at a solution
    i simplified it to

    6 CIS (pi/12)

    How do i turn it into cartesian?
     
  2. jcsd
  3. Mar 11, 2010 #2

    tiny-tim

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    Hi Stripe! :smile:

    (have a pi: π :wink:)

    I've never seen "CIS" before, but I'll guess you're right, and that it's cos + i*sin.

    Now use De Moivre's theorem … cosθ + isinθ = e :wink:

    And Cartesian form simply means in the form x + iy (as opposed to polar form, which is in the form re )

    (and no, it's not 6 CIS (π/12))
     
    Last edited: Mar 11, 2010
  4. Mar 11, 2010 #3

    HallsofIvy

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    "[itex]Cis(\theta)[/itex]" is engineering notation for "[itex]cos(\theta)+ i sin(\theta)[/itex]" which mathematicians tend to write as [itex]e^{i\theta}[/itex].

    The important thing about that notation is that [itex](Cis(\theta)*Cis(\phi)= Cis(\theta+ \phi)[/itex].

    So you have correctly deduced that [itex](2Cis(\pi/6))(3Cis(\pi/12)= 6 Cis(3\pi/12)= 6 Cis(\pi/4)[/itex]

    Now, just use the definition: [itex]6 Cis(\pi/4)= 6 cos(\pi/4)+ 6i sin(\pi/4)[/itex].
     
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