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Math Help - Analytic Functions?

  1. #1
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    Analytic Functions?

    Hello again,

    I am trying but it is really hard learning this stuff on your own. Never gonna take a calc course online again :\.

    This it he question, I am currently stuck on.

    Where is the function  f(z)=e^(^x^^^2^-^y^^^2^) * [cos(2xy) + isin(2xy)] analytic?

    For now could someone just help me with the derivative? I am in the process of figuring out Couchy-Rieman equations.

    Is the derivative only present if \partialu/ \partialx = \partialv/ \partialy?

    just for my understanding in the equation 1+iy there is no derivative because \partialu/ \partialx = 0 and \partialv/ \partialy = 1?

    Thanks
    Rioch
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  2. #2
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    Re: Analytic Functions?

    u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_x(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))
    v(x,y)=e^{x^2-y^2}\sin(2xy) \implies v'_y(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))
    u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_y(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))
    v(x,y)=e^{x^2-y^2}\sin(2xy) \implies -v'_x(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))
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  3. #3
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    Re: Analytic Functions?

    Quote Originally Posted by Ridley View Post
    u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_x(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))
    v(x,y)=e^{x^2-y^2}\sin(2xy) \implies v'_y(x,y)=2e^{x^2-y^2}(x\cos(2xy)-y\sin(2xy))
    u(x,y)=e^{x^2-y^2}\cos(2xy) \implies u'_y(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))
    v(x,y)=e^{x^2-y^2}\sin(2xy) \implies -v'_x(x,y)=-2e^{x^2-y^2}(x\cos(2xy)+y\sin(2xy))
    HI,

    Thanks for the quick response. I understand that you have taken the derivatives and in this example the Cauchy-Riemann equations are satisfied.

    I am unsure what this means though and where to go from here? No need to carry out the math (although it is helpful), however if possible a short explanation ?

    how would I express the final derivative? f'(x) , is it just the range of equations?
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  4. #4
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    Re: Analytic Functions?

    You asked where the function is analytic. The first step is to check where the partial derivatived exist and satisfy the Cauchy-Riemann equations, and as you can see they are satisfied everywhere. If the function is also continuous in that domain, then the function is analytic there (Looman-Menchoff's theorem).

    The derivative at that point can be written as \partial f/\partial z = u'_x+iv'_x = v'_y - iu'_y
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  5. #5
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    Re: Analytic Functions?

    Another approach:

    \begin{aligned} e^{(x^2-y^2)}(\cos(2xy) + i\sin(2xy)) &=  e^{(x^2-y^2)}e^{i2xy} \\ &= e^{x^2-y^2+2ixy} = e^{(x+iy)^2} = e^{z^2}. \end{aligned}

    Squaring and exponentiating are both analytic functions, so f(z) is analytic everywhere.
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