Integral of (z^2)dz around a semi-circle in the y >= 0 region.

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In summary, Simon was trying to integrate z^2 around the open semi-circle counter-clockwise from x=1 to x=-1, but the answer always comes out as 0. He was able to parameterize the line integral around the semi-circle using the cosine and sine of the angle between the radius vector and the semi-circle, but was not able to get the answer to match the size of the integral. When he tried integrate z around the circle using the eigenvalues of z, the answer always came out -4.
  • #1
Smin0
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Help! I'm preparing for Prelims (exams taken at the end of the first year of my course in Physics) and I can't do the line integral of (z^2)dz around the open semi-circle counter-clockwise from x=1 to x=-1. However I try to tackle it the answer comes out as 0, when I know it has some size because it cancels out with the straight line integral from -1 to 1 (possibly equal to 2/3, although I've lost the original working). The z is for a complex variable, but if anyone could at least just show me how to parameterize the regular line integral (not for complex, but just in 2D) around the semi-circle that would be so helpful. I've tried everyway I can (barring using cartesian, because I know there's a nice polar way to do it, I just keep coming out 0 when I try though), and it just won't work!

To re-iterate, the integral is (z^2)dz where z = x + iy.

Thank you!

Simon.
 
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  • #2
Smin0 said:
Help! I'm preparing for Prelims (exams taken at the end of the first year of my course in Physics) and I can't do the line integral of (z^2)dz around the open semi-circle counter-clockwise from x=1 to x=-1. However I try to tackle it the answer comes out as 0, when I know it has some size because it cancels out with the straight line integral from -1 to 1 (possibly equal to 2/3, although I've lost the original working). The z is for a complex variable, but if anyone could at least just show me how to parameterize the regular line integral (not for complex, but just in 2D) around the semi-circle that would be so helpful.
Yes, that's a perfectly valid way of doing it. Normally complex variables are written z= x+ iy where x and y are the real and imaginary parts. On the straight line from -1 to 1, z is always real so z= x looks like a perfectly good parameterization.

I've tried everyway I can (barring using cartesian, because I know there's a nice polar way to do it, I just keep coming out 0 when I try though), and it just won't work!

To re-iterate, the integral is (z^2)dz where z = x + iy.

Thank you!

Simon.
Unfortunately, you haven't show the work that "keeps coming out 0". On the unit circle, [itex]z= cos(\theta)+ i sin(\theta)[/itex] or equivalently but more simply, [itex]z= e^{i\theta}[/itex]. Now, how are you integrating that from 1 to -1? In particular, what limits of integration are you using?
 
  • #3
Int((z^2)dz). Also: my GOODNESS that latex took a long time to type out.

HallsofIvy said:
Unfortunately, you haven't show the work that "keeps coming out 0". On the unit circle, [itex]z= cos(\theta)+ i sin(\theta)[/itex] or equivalently but more simply, [itex]z= e^{i\theta}[/itex]. Now, how are you integrating that from 1 to -1? In particular, what limits of integration are you using?

Okay, well I thought I'd found the trouble when I realized that [tex]dz = izd\theta[/tex] instead of [tex]ird\theta[/tex], but alas it now comes out to -4. My working this time is as follows (r = 1):

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(cos\theta + isin\theta)^{2}(izd\theta)[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(cos\theta + isin\theta)^{2}(icos\theta - sin\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(icos^{3}\theta - isin^{2}\theta\cos\theta - 2sin\theta\cos^{2}\theta - cos^{2}\theta\sin\theta + sin^{3}\theta - i2sin^{2}\theta\cos\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(i(1 - sin^{2}\theta)cos\theta - isin^{2}\theta\cos\theta - 2sin\theta\cos^{2}\theta - cos^{2}\theta\sin\theta + (1 - cos^{2}\theta)sin\theta - i2sin^{2}\theta\cos\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(icos\theta - 3isin^{2}\theta\cos\theta - 3sin\theta\cos^{2}\theta + sin\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(sin\theta - 3sin\theta\cos^{2}\theta)d\theta + i\int^{\pi}_{0}(cos\theta - 3cos\theta\sin^{2}\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \left[- cos\theta - cos^{3}\theta\right]^{\pi}_{0} + i\left[sin\theta - 3sin^{3}\theta\right]^{\pi}_{0}[/tex]

[tex]\int{z}^{2}{dz} = ( - 1 - 1 - 1 - 1) + i(0 - 0 - 0 - 0)[/tex]

[tex]\int{z}^{2}{dz} = -4[/tex]

=/

What am I doing wrong here?
 
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  • #4
o___o

... okay, and when I do it:

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(e^{i\theta})^{2}(ie^{i\theta}d\theta)[/tex]

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(e^{i2\theta})(ie^{i\theta}d\theta)[/tex]

[tex]\int{z}^{2}{dz} = i\int^{\pi}_{0}(e^{i3\theta})d\theta[/tex]

[tex]\int{z}^{2}{dz} = i\left[\frac{1}{i3}e^{i3\theta}\right]^{\pi}_{0}[/tex]

[tex]\int{z}^{2}{dz} = \left[\frac{1}{3}e^{i3\theta}\right]^{\pi}_{0}[/tex]

[tex]\int{z}^{2}{dz} = \left(- \frac{1}{3} - \frac{1}{3}\right)[/tex]

[tex]\int{z}^{2}{dz} = - \frac{2}{3}[/tex]

which is what I believe to be the right answer ... how come it works this way but not the other?
 
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  • #5
Minus signs ...

Smin0 said:
[tex]\int{z}^{2}{dz} = ( - 1 - 1 - 1 - 1) + i(0 - 0 - 0 - 0)[/tex]

[tex]\int{z}^{2}{dz} = -4[/tex]

That should be [tex]-- 1 -- 1 -- 1 -- 1 = + 1 + 1 + 1 + 1 = 4[/tex] sorry. Still wrong though :(.

Oh, and:

[tex]\int{z}^{2}{dz} = \left[- cos\theta - cos^{3}\theta\right]^{\pi}_{0} + i\left[sin\theta - 3sin^{3}\theta\right]^{\pi}_{0}[/tex]

should be

[tex]\int{z}^{2}{dz} = \left[- cos\theta - cos^{3}\theta\right]^{\pi}_{0} + i\left[sin\theta - sin^{3}\theta\right]^{\pi}_{0}[/tex]
 
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  • #6
Just to be sure, this is contour you want to integrate z^2 on?

[tex]
\setlength{\unitlength}{2cm}
\begin{picture}(4,3)
\linethickness{0.4mm}
\qbezier(1,1)(1,2)(2,2)
\qbezier(2,2)(3,2)(3,1)
\linethickness{0.2mm}
\put(2.1,2.20){Imag}
\put(3.30,0.95){Real}
\put(2,0.75){\vector(0,1){1.5}}
\put(.75,1){\vector(1,0){2.5}}
\put(.835,.75){-1}
\put(2.95,.75){1}
\end{picture}
[/tex]

Lets call that contour "[tex]\gamma[/tex]", and also the same closed contour "[tex]\Gamma[/tex]".

Since z^2 has no poles we know that;

[tex]\oint_\Gamma z^2 \mathrm{d}z = 0[/tex]

And since we can break up [tex]\Gamma[/tex] into two parts ([tex]\gamma[/tex] and the line segment from -1 to 1), we get;

[tex]\oint_\Gamma z^2 \mathrm{d}z = \int_\gamma z^2 \mathrm{d}z + \int_{-1}^1 z^2 \mathrm{d}z = 0[/tex]

so

[tex]\int_\gamma z^2 \mathrm{d}z = -\int_{-1}^1 z^2 \mathrm{d}z[/tex]

And finally yes, [tex]\gamma[/tex] is parameterized (in [tex]\mathbb{C}[/tex]) by [tex]e^{it}[/tex] with t from 0 to pi. If you want to do it in [tex]\mathbb{R}^2[/tex] just take imaginary and real part.
 
  • #7
It looks like you just made some sort of simple algebra mistake in your first attempt with cos(q)+isin(q)
 
  • #8
Yep, that's the line integral. And I got the other part, from -1 to 1 along the x-axis, I just needed to learn to do the other part explicitly to prove that the closed contour sums to 0. I still can't see what's wrong with the integration in cosines and sines, but it comes out as the wrong answer - if anyone can see where I've gone wrong in it would be really helpful, I don't want to do the same thing in the exam when I can't check if it's right or not (so I can keep retrying it).

maze said:
It looks like you just made some sort of simple algebra mistake in your first attempt with cos(q)+isin(q)

I don't suppose you could see where though, could you? I got a friend over just to check it, and neither of us could see where the mistake is. I mean, I know there must be one, I don't think I've broken maths, but I don't know WHERE it is.

Thanks for everyone's help so far!
 
  • #9
As anticipated some algebraic mistakes :)

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(sin\theta - 4sin\theta\cos^{2}\theta)d\theta + i\int^{\pi}_{0}(cos\theta - 4cos\theta\sin^{2}\theta)d\theta[/tex]

[tex]\int{z}^{2}{dz} = \left[- cos\theta + \frac{4}{3}cos^{3}\theta\right]^{\pi}_{0} + i\left[sin\theta - \frac{4}{3}sin^{3}\theta\right]^{\pi}_{0}[/tex]

[tex]\int{z}^{2}{dz} = ( 1 - 4/3 + 1 - 4/3) [/tex]

[tex]\int{z}^{2}{dz} = -2/3[/tex]
 
  • #10
Thank you!

ansrivas said:
As anticipated some algebraic mistakes :)

[tex]\int{z}^{2}{dz} = \int^{\pi}_{0}(sin\theta - 4sin\theta\cos^{2}\theta)d\theta + i\int^{\pi}_{0}(cos\theta - 4cos\theta\sin^{2}\theta)d\theta[/tex]

Oh my goodness, I see where it happened now, it was the multiplying out of the [tex]i(1 - sin^{2}\theta)cos\theta[/tex], I just ignored the [tex]-isin^{2}\theta\cos\theta[/tex] term (and the similar [tex](1 - cos^{2})[/tex] bit). I spent forty minutes yesterday on this really simple check of the Cauchy-Riemann equations for complex differentiation because I'd missed a minus sign from when I'd crossed through the thing next to it and obscured it slightly, and so something I knew was diffable kept telling me it didn't satisfy them. I do like maths, but I wish I could do the algebra more consistently not wrong.

Thank you! At least I know I wasn't doing the method wrong now, just apparently can't multiply and add.

:)
 
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1. What is the formula for finding the integral of (z^2)dz around a semi-circle in the y >= 0 region?

The formula for finding the integral of (z^2)dz around a semi-circle in the y >= 0 region is ∫ (z^2)dz = ( z^3 / 3 ) + C, where C is the constant of integration.

2. How do you determine the boundaries of integration for this problem?

The boundaries of integration for this problem are determined by the limits of the semi-circle in the y >= 0 region. The starting point is the x-coordinate of the center of the circle, and the ending point is the x-coordinate of the point where the semi-circle intersects the x-axis.

3. Can you use the Fundamental Theorem of Calculus to solve this integral?

Yes, the Fundamental Theorem of Calculus can be used to solve this integral. This theorem states that the integral of a function can be found by evaluating the antiderivative of that function at the boundaries of integration.

4. Is the value of the integral affected by the radius of the semi-circle?

Yes, the value of the integral is affected by the radius of the semi-circle. As the radius increases, the value of the integral also increases.

5. Are there any special cases to consider when solving this integral?

One special case to consider when solving this integral is when the semi-circle is a complete circle. In this case, the integral becomes ∫ (z^2)dz = ( z^3 / 3 ) + C = ( R^3 / 3 ) + C, where R is the radius of the circle.

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