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Complex integer

  1. Sep 16, 2012 #1
    z=6*e2,5i

    Can anyone explain me ? The imaginary part = 3,59 and real part = -4,81

    I tried e(x) = cos x + i sin x, but it does not help me.
     
  2. jcsd
  3. Sep 16, 2012 #2

    CAF123

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    The complex number [itex] z [/itex] can be expressed in the form [itex] z = r(\cos(θ) + i\sin(\theta)), [/itex] where [itex] z = re^{i\theta}. [/itex] As long as theta is in radians, you should be able to read off the real and imaginary parts.
     
  4. Sep 16, 2012 #3

    Ray Vickson

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    If you mean eix = cos(x) + i*sin(x), then that certainly gives you the correct answer; here you need to use x = 2.5 (N. American style), or x = 2,5 (Euro style).

    Note, however, that the given answers are incorrect as exact statements; they are only approximations to the true values, which are approximately
    real part ≈ -4.806861693281602289001016742804109986572
    and
    im part ≈ 3.590832864623738964311128213116973630222
    to 40-digit accuracy. No matter how many digits we use we will never be able to write down the exact value.

    RGV
     
  5. Sep 16, 2012 #4
    How do I calcuate the values if x=2,5 ?
     
  6. Sep 16, 2012 #5

    Ray Vickson

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    What is stopping you from calculating cos(2,5) and sin(2,5)? (Remember, though, that the '2,5' is in radians, not degrees.)

    RGV
     
  7. Sep 16, 2012 #6
    okey tnx, i undrestood, but i have one other question that if there is no number in the middle for example z=e-5i then r = 1 ?
     
  8. Sep 17, 2012 #7

    CAF123

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    Yes.
     
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