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Math Help - CAUCHY - RIEMANN Help please

  1. #1
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    CAUCHY - RIEMANN Help please

    Why did Cauchy - RIemann invented this.. arrghh... help


    Show that sin(z) satisfies the CAUCHY - RIEMANN conditions for analyticity for all values of z. Does [1/sin(z)] satisfy similar conditions? Calculate the derivative of [1/sin(z)] at z=0, . Comment on each answer.
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    Re: CAUCHY - RIEMANN Help please

    Have you tried going through the Cauchy-Riemann conditions with this function? If not, you should give it a go!
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by timeforchg View Post
    Why did Cauchy - RIemann invented this.. arrghh... help
    So that you know when a complex function is differentiable...

    Have you started by writing \displaystyle \begin{align*} \sin{(z)} = \sin{(x + i\,y)} = \sin{x}\cosh{y} + i \cos{x}\sinh{y} = u(x, y) + i\,v(x, y) \end{align*} and evaluating their partial derivatives?
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    Re: CAUCHY - RIEMANN Help please

    f(z) = sin (z)
    = sin (x + iy)
    = sin x cosh y + i cos x sinh y

    thus,

    u(x,y)=sin x cosh y ....... v(x,y)= cos x sinh y

    du/dx = cos x ............ dv/dx = -sin x
    du/dy = -sinh y ........... dv/dy = cosh y

    therefore I will compare it with the Cauchy Riemann formula.

    Am I right to say it doesn't satisfies the condition??
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by timeforchg View Post
    f(z) = sin (z)
    = sin (x + iy)
    = sin x cosh y + i cos x sinh y

    thus,

    u(x,y)=sin x cosh y ....... v(x,y)= cos x sinh y

    du/dx = cos x ............ dv/dx = -sin x
    du/dy = -sinh y ........... dv/dy = cosh y

    therefore I will compare it with the Cauchy Riemann formula.

    Am I right to say it doesn't satisfies the condition??
    Your calculation of the partial derivatives is incorrect. When calculating a partial derivative, all other variables are held constant, so are treated as constants. So with \displaystyle \begin{align*} u = \sin{x}\cosh{y} \end{align*}, when evaluating \displaystyle \begin{align*} \frac{\partial u}{\partial x} \end{align*}, the \displaystyle \begin{align*} \cosh{y} \end{align*} is treated as constant, and constant multiples in derivatives stay where they are... So it should be \displaystyle \begin{align*} \frac{\partial u}{\partial x} = \cos{x}\cosh{y} \end{align*}. Fix the rest.
    Thanks from timeforchg
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    Re: CAUCHY - RIEMANN Help please

    thanks for notifying me the error and yes they satisfy the condition.
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    Re: CAUCHY - RIEMANN Help please

    If I were to find whether 1/sin(z) satisfy the C.R condition,
    Do I use 1/sin(x) cosh(y) + j cos(x) sinh(y) or the one with inverse [sin(x) cosh(y) + j cos(x) sinh(y)]^-1.
    How do I change it to (x+iy)?
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by timeforchg View Post
    If I were to find whether 1/sin(z) satisfy the C.R condition,
    Do I use 1/sin(x) cosh(y) + j cos(x) sinh(y) or the one with inverse [sin(x) cosh(y) + j cos(x) sinh(y)]^-1.
    How do I change it to (x+iy)?
    Multiply top and bottom by the bottom's conjugate.
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by Prove It View Post
    Multiply top and bottom by the bottom's conjugate.
    when i try to conjugate. I get 0. so does it meant it does not exist?
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    Re: CAUCHY - RIEMANN Help please

    If the denominator is \displaystyle \begin{align*} \sin{x}\cosh{y} + j\cos{x}\sinh{y}  \end{align*}, the conjugate is \displaystyle \begin{align*} \sin{x}\cosh{y} - j\cos{x}\sinh{y} \end{align*}. Multiplying them together will give \displaystyle \begin{align*} \sin^2{x}\cosh^2{y} + \cos^2{x}\sinh^2{y} \end{align*}. This is not 0.
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by Prove It View Post
    If the denominator is \displaystyle \begin{align*} \sin{x}\cosh{y} + j\cos{x}\sinh{y}  \end{align*}, the conjugate is \displaystyle \begin{align*} \sin{x}\cosh{y} - j\cos{x}\sinh{y} \end{align*}. Multiplying them together will give \displaystyle \begin{align*} \sin^2{x}\cosh^2{y} + \cos^2{x}\sinh^2{y} \end{align*}. This is not 0.
    I use the wrong equation that is 1\sin(x+iy) thus I conjugate it. Oops. Once again thanks. will try it now.
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    Re: CAUCHY - RIEMANN Help please

    Yes your right. I did not get 0 when I do it manually. By using my calculator my answer will get 0.



    btw am i right to say that 1/sin (z) in (a+jb) format is

    sin(x) cosh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) - j cos(x) sinh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) ??

    Correct me if I'm wrong
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    Re: CAUCHY - RIEMANN Help please

    Quote Originally Posted by timeforchg View Post
    Yes your right. I did not get 0 when I do it manually. By using my calculator my answer will get 0.



    btw am i right to say that 1/sin (z) in (a+jb) format is

    sin(x) cosh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) - j cos(x) sinh(y)/ sin^2 (x) cosh^2 (y) + cos^2 (x) sinh^2 (y) ??

    Correct me if I'm wrong
    It's nasty to do it manually..

    If by theorem, 1/sin(z) will satisfy Cauchy - Riemann conditions for all values of z except z = k pi + pi/2, ( k=0, +- 1, +- 2 ...), where the denominator of the function equals to zero. ??


    but I try it out, it seems it doesn't satisfy the equation. hmm...
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    Re: CAUCHY - RIEMANN Help please

    SOLVED!! Thanks a lot guys!! Appreciate it.
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