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Math Help - Complex Variables

  1. #1
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    Complex Variables

    consider the complex function

    f(z) = e ^(2z^(2)+2)

    let f(x+iy)=u(x,y)+iv(x,y)
    calculate u(x,y) and v(x,y)

    after substituing z=x+iy I get

    e^(2x^(2)-2y^(2)+2) . e^(4xiy)

    how do i get rid of e.
    if i have done it wrong then plz tell me whats the mistake.plz help
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  2. #2
    Member ronaldo_07's Avatar
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    to change it to e I think you change it in the form of e^itheta = cos(theta)+isin(theta)
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  3. #3
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    so what do u get as u and v
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  4. #4
    Member ronaldo_07's Avatar
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    Im not sure myself how to follow on, Im hoping someone else could help.
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  5. #5
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    i have got it.lemme help u ronaldo

    f(z) = e^(2z + 2)

    f(x + iy)
    = e^(2(x + iy) + 2)
    = e^(2(x + 2ixy + iy) + 2)
    = e^(2(x + 2ixy - y) + 2)
    = e^(2x + 4ixy - 2y + 2)
    = e^(4ixy)e^(2x - 2y + 2)
    = [cos(4xy) + isin(4xy)]e^(2x - 2y + 2)
    = cos(4xy)e^(2x - 2y + 2) + isin(4xy)e^(2x - 2y + 2)

    So:
    u(x,y) = cos(4xy)e^(2x - 2y + 2)
    v(x,y) = sin(4xy)e^(2x - 2y + 2)
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  6. #6
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    is z^2=x^2+y^2 or x^2-y^2?
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  7. #7
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    Quote Originally Posted by usman206 View Post
    consider the complex function f(z) = e ^(2z^(2)+2)
    You notation is very hard to follow. If this is it, then
    \begin{gathered}<br />
  2z^2  + 2 = \left( {2x^2  - 2y^2  + 2} \right) + i\left( {4xy} \right) \hfill \\<br />
  e^{2z^2  + 2}  = e^{\left( {2x^2  - 2y^2  + 2} \right)} \left[ {\cos (4xy) + i\sin (4xy)} \right] \hfill \\ <br />
\end{gathered}
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