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Finding the polar form of a complex number

  1. Mar 5, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-3-5_22-32-46.png
    2. Relevant equations
    r=sqrt(a^2+b^2)
    θ=arg(z)
    tan(θ)=b/a

    3. The attempt at a solution
    for a)

    upload_2017-3-5_22-36-19.png
    finding the polar form:
    r=sqrt(-3^2+(-4)^2)=sqrt(7)
    θ=arg(z)
    tan(θ)=-4/-3 = 53.13 °
    300-53.13=306.87°

    -3-j4=sqrt(7)*(cos(306.87+j306.87)

    I don't know if my answer is correct because it is given that -π<arg(z)<=π
     

    Attached Files:

  2. jcsd
  3. Mar 5, 2017 #2

    BvU

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    Is indeed the wrong answer. The second = sign makes no sense. You probably mean to say $$ \tan\theta = {-4\over -3}\ \ \Rightarrow \ \ \theta = 53.13^\circ$$but that is not correct. To satisfy
    and to not fold everything back to the range ##(-\pi/2 , \pi/2]## the atan2 function was 'invented'.
    Your 300-53.13=306.87° (an alternative answer ?) doesn't satisfy "-π<arg(z)<=π, not even if converted to radians.

    And to top it all up, ##3^2+4^2 \ne 7## !

    So it's back to the drawing board, I'm afraid...:wideeyed:

    Not much help (except the reference to atan2, perhaps), but with your drawing at hand I'm pretty confident you can manage on your own and that's much better.
     
  4. Mar 5, 2017 #3
    Yes, that was what I meant.

    my bad its equal 25, meaning r = 5.

    I will try to read about atan2, to be honest I didn't knew about it. Thank you
     
  5. Mar 6, 2017 #4

    Ray Vickson

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    (1) Why are you using degrees when the question expresses angles in radians?
    (2) In your computation 300-53.13, where does the 300 come from?
     
  6. Mar 6, 2017 #5

    Mark44

    Staff: Mentor

    Pretty clearly, it's a typo, where the OP typed 300, but meant 360.
     
  7. Mar 6, 2017 #6

    Ray Vickson

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    OK, then: but where does the 360 come from?
     
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