My Math Forum Integral on a circumference

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 November 1st, 2016, 04:30 AM #1 Newbie   Joined: Oct 2016 From: italy Posts: 9 Thanks: 0 Integral on a circumference Hi guys, I got this function: $\displaystyle g(t) = \frac{e^{iz}}{(z)^{1/3}}$ I have to integrate it's derivate $\displaystyle g'(t)$ on the circumference of center $\displaystyle 0$ and radius $\displaystyle \pi$ oriented counterclockwise. The problem says also that $\displaystyle arg(z)$ is between $\displaystyle ]-\pi,\pi[$ (determination of the cubic root). I thought to integrate the derivate and to apply Cauchy (the integral should be equal to the sum of the residues, but the residue in 0 is 0). Suggestions? Last edited by nostradamus1915; November 1st, 2016 at 04:33 AM.
 November 1st, 2016, 05:50 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra OK, so you have the complex plane with a branch cut from zero out along the negative real axis. The integral of the derivative exists everywhere else, so the integral along any path that does not cut the negative real axis is $$\int_C f'(z)\,\mathrm dz = f(b)-f(a)$$ where $C$ is a curve from $z=a$ to $z=b$. Now, to evaluate your integral, pick 2 points $z_1,z_2$ either sides of the branch cut on the given circle and let z them move towards the branch cut. Take the limit of the expression at the branch cut where the denominator is expressed in the form $z=re^{i\theta}$. Thanks from topsquark, Country Boy and nostradamus1915
November 6th, 2016, 09:04 AM   #3
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Quote:
 Originally Posted by v8archie Now, to evaluate your integral, pick 2 points $z_1,z_2$ either sides of the branch cut on the given circle and let z them move towards the branch cut. Take the limit of the expression at the branch cut where the denominator is expressed in the form $z=re^{i\theta}$.
I don't fully understand how to do that, may you show me please? Besides, there is no result on this exercise, so I'm not sure if I did good or not.

 November 6th, 2016, 10:14 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra The branch cut for the cube root is along the negative real axis because we are given $-\pi \lt \arg z \lt \pi$. The derivative $g'(t)$ therefore exists everywhere except on the negative real axis (where $arg z$ has a jump discontinuity) and $z=0$ where $g(z)$ does not exist. The integral of $g'(z)$ therefore exists everywhere else and is equal to $g(z)$. Therefore, the integral along any path from point $z_1$ to point $z_2$ that avoids zero and the negative real axis is equal to $g(z_2)-g(z_1)$. Thus, we $z_1$ below the negative real axis on the circle $|z|=\pi$ and $z_2$ on the same circle but above the real axis. We define our path to be the part of the circle from $z_1$ to $z_2$ in an anticlockwise direction which avoids zero and the negative real axis, so the integral is $g(z_2)-g(z_1)$. Now, we let $z_1$ and $z_2$ move around the circle towards the negative real axis, completing the circle. In this way $z_1 \to \pi e^{-\pi}=-\pi$ and $z_2 \to \pi e^{\pi}=-\pi$. Thus $$g(z_2)-g(z_1) \to \frac{e^{-i\pi}}{ \pi^{\frac13} e^{i\frac{\pi}3}}-\frac{e^{-i\pi}}{ \pi^{\frac13} e^{i\frac{-\pi}3}}$$ Thanks from nostradamus1915 Last edited by v8archie; November 6th, 2016 at 11:14 AM.
 November 7th, 2016, 10:56 AM #5 Newbie   Joined: Oct 2016 From: italy Posts: 9 Thanks: 0 Thanks for the help, just last doubt, shouldn't $\displaystyle g(z_1)$'s numerator be $\displaystyle e^{i \pi}$ ?
 November 7th, 2016, 02:33 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra No, because both $z_1$ and $z_2$ are heading towards $-\pi + 0i$ but with arguments that are $2\pi$ different.
 November 8th, 2016, 03:18 AM #7 Newbie   Joined: Oct 2016 From: italy Posts: 9 Thanks: 0 But I don't get why the denominators are different, if $\displaystyle z=-\pi$ in both cases, shouldn't be $\displaystyle e^{\frac{i \pi}{3}}$ for both $\displaystyle g(z_1)$ and $\displaystyle g(z_2)$?
November 8th, 2016, 03:45 AM   #8
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Quote:
 Originally Posted by nostradamus1915 But I don't get why the denominators are different, if $\displaystyle z=-\pi$ in both cases, shouldn't be $\displaystyle e^{\frac{i \pi}{3}}$ for both $\displaystyle g(z_1)$ and $\displaystyle g(z_2)$?
No, because in the denominator we are using the representation $z=|z|e^{i\arg z}$. If you use that representation in the denominators you get
$$e^{-iz_1}=e^{-i|z_1|e^{i\arg z_1}} = e^{-i\pi e^{-i\pi}} = e^{-i\pi (-1)} = e^{i\pi}=-1$$
Similarly
$$e^{-iz_2}=e^{-i|z_2|e^{i\arg z_2}} = e^{-i\pi e^{i\pi}} = e^{-i\pi (-1)} = e^{i\pi}=-1$$

Last edited by v8archie; November 8th, 2016 at 03:50 AM.

 November 8th, 2016, 05:28 AM #9 Global Moderator   Joined: Dec 2006 Posts: 17,172 Thanks: 1285 In your earlier post, you omitted $i$ from two exponents.
 November 8th, 2016, 09:46 AM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra Thanks.

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