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Homework Help: Finding polar form of complex number

  1. Nov 19, 2015 #1
    1. The problem statement, all variables and given/known data
    I have the following complex numbers : -3,18 +4,19i
    I must put it in polar form.

    2. Relevant equations
    r=(a^2+b^2)^(1/2)
    cos x = a/r
    sin x = b/r

    3. The attempt at a solution

    I was able to find with cos x = a/r that the x = 127,20

    But when I do it with sin x = b/r I obtain like 52 degrees. I know that I Must obtain 127,20 for BOTH. Why isnt it working ?
     
  2. jcsd
  3. Nov 19, 2015 #2

    FactChecker

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    The sin of both those angles are the same, so you must decide which is correct. That is the angle that has the correct a,b values. 52 degrees would be at (3.18, 4.19) and 127.2 degrees is at (-3.18, 4.19).
     
  4. Nov 19, 2015 #3
    Oh ok so its normal that I obtain two different values. Ok then, thank you!
     
  5. Nov 20, 2015 #4

    HallsofIvy

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    Yes, There are always two values of [itex]\theta[/itex] in the interval 0 to [itex]2\pi[/itex] that have the same [itex]sin(\theta)[/itex]. But you still have to determine which is correct for the specific problem- the two different values, [itex]\theta[/itex] have different values for [itex]cos(\theta)[/itex].
     
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