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Math Help - Complex function

  1. #1
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    Complex function

    Show that

    f(z) = \left\{\begin{array}{ll} \cos (z^{1/2}),& z\neq 0\\1,&z=0\end{array}

    is single-valued and analytic in C. Calculate f'(0).

    I don't even know where to begin with this one.
    Last edited by Mondreus; June 22nd 2010 at 01:15 AM.
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  2. #2
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    I don't understand why this is listed as a hybrid function, since \cos{0} = 1 anyway... So I'm going to assume we can just write the function as f(z) = \cos{(z^{\frac{1}{2}})} for all z.


    f(z) = \cos{(z^{\frac{1}{2}})}

     = \cos{[(r\,e^{i\theta})^{\frac{1}{2}}]}

     = \cos{(r^{\frac{1}{2}}e^{\frac{\theta}{2}i})}

     = \cos{\left[r^{\frac{1}{2}}\left(\cos{\frac{\theta}{2}} + i\sin{\frac{\theta}{2}}\right)\right]}

     = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}} + i\,r^{\frac{1}{2}}\sin{\frac{\theta}{2}}\right)}

    = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r  ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta  }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\  right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet  a}{2}}\right)}


    Now convert it back to Cartesians and use the Cauchy Riemann Equations.

    i.e. For f(z) = u(x, y) + i\,v(x, y), the function is analytic if \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} and \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
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  3. #3
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    The first order partial derivatives must be continuous as well in order for the function to be analytic, which I'm guessing is a part of the problem. The reason the function looks like that probably has something to do with z^{1/2} = e^{\frac{1}{2}log(z)} when z \neq 0.
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  4. #4
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    Well, we don't know exactly what you are allowed to use - can you use the fact that the cosine function is entire and square root is analytic in  \mathbb{C} - \{ 0 \}?

    Also, it is not defined at 0 since, as you said,  \sqrt{z} is defined as e^{\frac{1}{2} log(z)}, and logz has a branch point at 0.

    I don't think it is 1:1, though -- f(4 \pi ^2) = f(0). Regarding the calculation of the derivative - you should probably do it by definition.
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  5. #5
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    One-to-one might be the wrong wording here. You're supposed to show that it's not multivalued. You're allowed to use that cos is entire, but like I wrote previously, I think you need to show that f and the partial derivatives are continuous. Anyways, this is just an exercise, but I would really like to know how to solve it to further my knowledge.
    Last edited by Mondreus; June 22nd 2010 at 02:34 AM.
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  6. #6
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    Actually, the Cauchy-Riemann Equations in Polar Coordinates is more useful in this case.

    You need to show that \frac{\partial u}{\partial r} = \frac{1}{r}\,\frac{\partial v}{\partial \theta} and \frac{\partial v}{\partial r} = -\frac{1}{r}\,\frac{\partial u}{\partial \theta}.


    So for your function...

    f(z) = \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r  ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta  }{2}}\right)} - i\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\  right)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\thet  a}{2}}\right)}


    u =  \cos{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r  ight)}\cosh{\left(r^{\frac{1}{2}}\sin{\frac{\theta  }{2}}\right)} and v = -\sin{\left(r^{\frac{1}{2}}\cos{\frac{\theta}{2}}\r  ight)}\sinh{\left(r^{\frac{1}{2}}\sin{\frac{\theta  }{2}}\right)}.

    Can you go from here?
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  7. #7
    MHF Contributor chisigma's Avatar
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    Remembering the Taylor expansion of cos(*) You can write...

    \displaystyle \cos \sqrt{z} = \sum_{n=0}^{\infty} (-1)^{n} \frac{z^{n}}{(2n)!} (1)

    ... so that \cos \sqrt{z} is analytic on the whole z plane. Its derivative can be computed deriving (1) 'term by term' and is f' (0)= - \frac{1}{2}...

    Kind regards

    \chi \sigma
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  8. #8
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    The Cauchy-Riemann equations is only a necessary condition in order for the function to be analytic. In order to prove that the function is analytic you also need to show that the partial derivatives are continuous. I don't know how to get the partial derivatives for the function though since it's a split function. I would think that the derivative of the constant part is 0, but the answer is f'(0)=-1/2
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  9. #9
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    To be continuous at that point, you just need to show that

    \lim_{z \to 0 + 0i}f(z) = 1.
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