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Math Help - show the function is an entire function

  1. #1
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    show the function is an entire function

    Hi, can someone please help me..
    Question: Write the function f(z) = sin2(z) in the form  f = u + iv , with u and v real functions. Hence show that f(z) is an entire function.

    here's my working:

    <br />
f(z) = sin2(z), where z = x + iy
     = sin2(x + iy) <br />
= sin 2x cos 2iy + cos 2x sin iy

    since,
     f(z) = u + iv
    thus  u(x,y) = 0, v (x,y) = sin 2x cos 2iy + cos 2x sin iy

    then i do partial derivaties to satisfy cauchy riemann equations..
    partial of u wrt x = 0
    partial of v wrt y = cos 2x i sin iy - 2i sin 2x cos 2iy
    partial of u wrt y = 0
    partial of v wrt x = ..

    well i know i did it wrong coz the following are meant to be satisfied, but they just don't work out:
    partial of u wrt x = partial of v wrt y
    and
    partial of u wrt y = - partial of v wrt x

    so can someone please help me??
    Last edited by Dgphru; March 24th 2010 at 11:46 PM.
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  2. #2
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    Quote Originally Posted by Dgphru View Post
    Hi, can someone please help me..
    Question: Write the function f(z) = sin2(z) in the form  f = u + iv , with u and v real functions. Hence show that f(z) is an entire function.

    here's my working:

    <br />
f(z) = sin2(z), where z = x + iy
     = sin2(x + iy) <br />
= sin 2x cos 2iy + cos 2x sin iy

    since,
     f(z) = u + iv
    thus  u(x,y) = 0, v (x,y) = sin 2x cos 2iy + cos 2x sin iy

    then i do partial derivaties to satisfy cauchy riemann equations..
    partial of u wrt x = 0
    partial of v wrt y = cos 2x i sin iy - 2i sin 2x cos 2iy
    partial of u wrt y = 0
    partial of v wrt x = ..

    well i know i did it wrong coz
    partial of u wrt x = partial of v wrt y
    and
    partial of u wrt y = - partial of v wrt x

    so can someone please help me??
    \sin{2z} = \sin{2(x + iy)}

     = \sin{(2x + 2iy)}

     = \sin{2x}\cos{2iy} + \cos{2x}\sin{2iy}


    You should know that \sin{iX} = i\sinh{X} and \cos{iX} = \cosh{X}.

    So the equation becomes

     = \sin{2x}\cosh{2y} + i\cos{2x}\sinh{2y}

     = u + iv.


    For the function to be an entire function, the Cauchy-Riemann Equations need to be satisfied.

    i.e. you need \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} and \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}.


    u = \sin{2x}\cosh{2y}

    \frac{\partial u}{\partial x} = 2\cos{2x}\cosh{2y}

    \frac{\partial u}{\partial y} = 2\sin{2x}\sinh{2y}


    v = \cos{2x}\sinh{2y}

    \frac{\partial v}{\partial x} = -2\sin{2x}\sinh{2y}

     = -\frac{\partial u}{\partial y} as required.

    \frac{\partial v}{\partial y} = 2\cos{2x}\cosh{2y}

     = \frac{\partial u}{\partial x} as required.


    Since the function satisfies the Cauchy-Riemann Equations for all x, y, the function is an entire function.
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