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Polar to cartesian form

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Convert 2cis(-pi/3)cis(pi/6) into cartesian form. Show all working to obtain full marks

    2. Relevant equations

    I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))

    3. The attempt at a solution

    Okay so cos of (-p/3) = 1/2
    Sin of (-p/3) = sqrt(3)/2
    Cos(pi/6)=sqrt(3)/2
    Sin(pi/6) = 1/2

    So basically you get this big answer as 2((1/2-isqrt(3)/2)+(sort(3)/2+1/2i))

    Is this the correct answer or do I need to multiply it out?

    I know the R for this is 2. I know the theta is tan^-1(y/x) but I am a little confused...
     
  2. jcsd
  3. Oct 21, 2014 #2

    Mark44

    Staff: Mentor

    In the problem statement you're multiplying, but in the above you're adding. Which is it?
    The line above is wrong.
     
    Last edited: Oct 21, 2014
  4. Oct 21, 2014 #3
    The reason I am adding is because is that not the formula to convert from that form to Cartesian?


    What is wrong?
     
  5. Oct 21, 2014 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    ##\text{cis}(a) \times \text{cis}(b) = \text{cis}(a+b)##; see, eg.,http://math.wikia.com/wiki/Cis_θ
     
  6. Oct 21, 2014 #5

    Mark44

    Staff: Mentor

    This is what you wrote: "I know that the equation for it is 2((cos(theta) +isin(theta))+(cos(theta)+isin(theta)))"
    That is incorrect, and isn't what the formula says.
    You are misunderstanding what the formula says.
     
  7. Oct 21, 2014 #6

    BvU

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    Science Advisor
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    Gold Member

    The instruction "show all working" leads me to believe they want you to write out the whole thing, thus 'discovering' the wikia product theorem.
    Don't work with numbers, just leave the ##-{\pi\over 3}## and the ##\pi \over 6## until the very last.
     
  8. Oct 27, 2014 #7
    Thanks for that guys, I think I got it now!!!
     
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