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Find the real and imaginary part of sin(4+3i)

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the real and imaginary part of sin(4+3i)

    2. Relevant equations

    sinx = [tex]\frac{e^z - e^(-z)}{2i}[/tex]

    cosx = [tex]\frac{e^z + e^(-z)}{2}[/tex]

    sin(iy) = i[tex]\frac{e^y - e^(-y)}{2}[/tex]

    cos(iy) = [tex]\frac{e^y + e^(-y)}{2}[/tex]

    various trig identities

    3. The attempt at a solution

    So I used sin(x+y) trig identity and got
    sin4*cos3i + sin3i*cos4

    I turned them all into exponents using the appropriate equations stated in (2).

    I got to a point where nothing is really calculable by hand/head. Is there an easier way to do this or does the calculator need to be used at a certain point to calculate the real part(terms grouped w/o i) and the imaginary part (terms grouped with i).

    If so, then I guess I need help getting the terms grouped together to calculate the real and imaginary parts.

    Where I am stuck is at:

    [tex]\frac{e^{3+4i}+e^{-3+4i}-e^{3-4i}+e^{-3-4i}}{4i}[/tex] +
    [tex]\frac{e^{3+4i}-e^{-3+4i}+e^{3-4i}-e^{-3-4i}}{4}[/tex]

    (the two fractions should be added together)

    Now what should I do with all these lovely exponents? Should I have even gone this route?
     
  2. jcsd
  3. Sep 18, 2010 #2

    HallsofIvy

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    Now use [itex]e^{3+ 4i}= e^3cos(4)+ i e^3sin(4)[/itex], etc.
     
  4. Sep 18, 2010 #3
    That was just the hint I needed, Halls. Thanks! Finally got it.
     
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