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Exponential complex number question

  1. Mar 14, 2010 #1
    Find the real part of the complex number:

    [tex](1-12i)e^{-2+4i}[/tex]

    I know that z = a + ib can be rewritten as

    [tex]z = |z|e^{i\theta}[/tex]

    but that doesn't help because the coefficient of the e is not a scalar value, rather a complex number.

    Thanks
    Tom
     
  2. jcsd
  3. Mar 14, 2010 #2

    tiny-tim

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    Homework Helper

    Hi Tom! :smile:

    Just write e-2+4i in Cartesian form (ie a + ib), and multiply the whole thing out. :wink:
     
  4. Mar 14, 2010 #3
    So,

    z = a + ib

    [tex]
    (1-12i)e^{-2+4i}
    [/tex]

    [tex]
    => (1-12i)e^{-2}e^{4i}
    [/tex]

    we can say [tex]
    e^{4i} = (cos(4^{c})+isin(4^{c}))
    [/tex]

    all good so far?

    so

    [tex]
    => (1-12i)e^{-2}(cos(4^{c})+isin(4^{c}))
    [/tex]

    we can write the second part of the term in Cartesian (or first part in polar):

    [tex](1-12i)\cdot e^{-2}\cdot(\frac{1+4}{\sqrt{17}})[/tex]

    Something is not right as the angle part (b/a) can be ambiguous - 8/2, 4/1, 16/4 ect

    Is that the correct answer?

    Thanks
    Thomas
     
  5. Mar 14, 2010 #4
    not sure if this will help:

    setting up simultaneous equations:
    [tex]1 = \sqrt(b^{2} + a^{2})[/tex]
    [tex]1 = b^{2} + a^{2}[/tex]

    and also

    [tex]tan\theta = \frac{b}{a}[/tex]

    so [tex]b = atan\theta[/tex]

    and [tex]b^{2} = 1 - a^{2}tan\theta[/tex]

    That any good?
     
  6. Mar 14, 2010 #5

    tiny-tim

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    Yes (but what's c ? … 4 is in radians).

    Now just multiply it out, and only use the bits of the result that don't have i. :smile:
     
  7. Mar 14, 2010 #6
    Genius :)
     
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