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Math Help - complex variables

  1. #1
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    complex variables

    How do I express sin(e^i) in terms of z = x + yi with x,y in R.

    I say that e^i = cos1 + i*sin1 (since e^z = e^x(cosy + i*siny).
    Then sin(cos1 + i*sin1) = [e^(i*(cos1 + i*sin1)) - e^(-i*(cos1 + i*sin1))]/2i, which equals (e^(i*cos1 - sin1) - e^(-i*cos1 + sin1))/2i. How do I write this in the form x + yi?

    Thanks for your help in advance.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PvtBillPilgrim View Post
    How do I express sin(e^i) in terms of z = x + yi with x,y in R.

    I say that e^i = cos1 + i*sin1 (since e^z = e^x(cosy + i*siny).
    Then sin(cos1 + i*sin1) = [e^(i*(cos1 + i*sin1)) - e^(-i*(cos1 + i*sin1))]/2i, which equals (e^(i*cos1 - sin1) - e^(-i*cos1 + sin1))/2i. How do I write this in the form x + yi?

    Thanks for your help in advance.
    note that if a complex number is given in the form re^{i \theta} then we can find its rectangular form x + iy using the following relationships:

    \cos \theta = \frac xr \implies \boxed{x = r \cos \theta} and \sin \theta = \frac yr \implies \boxed{y = r \sin \theta}
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  3. #3
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    Quote Originally Posted by PvtBillPilgrim View Post
    How do I express sin(e^i) in terms of z = x + yi with x,y in R.
    Here are two that may help you.
    \sin (z) = \frac{1}{{2i}}\left( {e^{iz}  - e^{ - iz} } \right)

    \sin (x + yi) = \sin (x)\cosh (y) + i\left[ {\cos (x)\sinh (y)} \right]
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