Fundamental Theorem of calculus & Estimation Theorem

**moolimanj** · June 14th 2007, 12:09 AM

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?

**ThePerfectHacker** · June 14th 2007, 06:33 AM

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.

**ThePerfectHacker** · June 14th 2007, 06:35 AM

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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