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Math Help - Fundamental Theorem of calculus & Estimation Theorem

  1. #1
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    Fundamental Theorem of calculus & Estimation Theorem

    Hi Guys

    Can you help with the attached problem? I think for part (i) and (ii) the answer is the same - i have this as 0 in both cases. However, i'm having problems with part (a) (iii). How do i prove that its not an entire function. Has this got anything to do with the Grid Path Theorem?

    For part (b) i have an answer of (3pi/26)(9+e^6). For some reason, this looks incorrect. Can someone work through the problem and tell me if this is what they also get?

    For part (c) (i) I get a final answer of -e^3 + e^-3. Is this right and if so does it simplify into a trigonometric function?
    For part (c) (ii) I get a final answer of 0. I have done the following for this:

    f(z)=zcosh(z^2) = 0.5(2z)cosh(z^2). Hence a primitive of f is f(z) = 0.5sinh(z^2) and the values of 3 and -3 are input into this and subtracted from each other to give 0. Is this correct?
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  2. #2
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    1)If f(z) = 2\bar z then f(2it) = 2\bar{(2it)} = 2(-2it)=-4it and \gamma_1 ' (t) = 2i. That means \int_{\gamma_1} 2\bar z dz = \int_{-1}^1 (-4it)(2i) dt = \int_{-1}^1 8t dt = 4t^2\big|_{-1}^1 = 0.

    Now if \gamma_2(t) = 2e^{it} then f(2e^{it}) = f(2\cos t + 2i\sin t) = 4\cos t - 4i\sin t. With \gamma_2'(t) = -2\cos t + 2i\cos t. Then \int_{\gamma_2} 2 \bar z dz = \int{-\pi/2}^{\pi/2} (4\cos t - 4i\sin t)(-2\sin t + 2i\cos t) dt \not = 0

    The problem if that if there existed an entire function g such that g'(z) = 2\bar z then the two integrals must be the same because the Fundamental Theorem of Line Integrals implies path independence.
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  3. #3
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    3)Note that \left( -e^{-z} \right)' = e^{-z}. And \left( \frac{1}{2}\sinh z^2 \right)' = z\cosh z^2. The rest is trivial.
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