- #1

- 8

- 0

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

Thank you!

Simon.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Smin0
- Start date

- #1

- 8

- 0

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

Thank you!

Simon.

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 969

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.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.

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,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.

- #3

- 8

- 0

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,howare 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 realised 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?

Last edited:

- #4

- 8

- 0

... 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?

Last edited:

- #5

- 8

- 0

[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]

Last edited:

- #6

- 107

- 0

[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

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

- 659

- 4

- #8

- 8

- 0

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

- 26

- 0

[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

- 8

- 0

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.

:)

Last edited:

Share: