Contour Integration

Where have the 'z's in the numerator of your last line of maths come from? (I may well just be being stupid and/or need to do some actual calculations, but I can't see it).

Edit: (You'll also find life easier if you rewrite

(z+i\sqrt{3})(z-i\sqrt{3})

as z^2 + 3).

Reply 2

Original post by Perpetuallity

And now I need to apply the cauchy integration formula, but I don't know where to begin! :^_^:

I don't understand where you're stuck. It should be fairly obvious how to proceed from there. How well do you know Cauchy's integral formula? Notice that your poles are both of first order and since the singularities are at

z=\sqrt{\dfrac{1}{3}}i

and

z=-\sqrt{\dfrac{1}{3}}i

respectively for each term.

It follows that:

- \frac{4}{3}i\left( \displaystyle\oint_{C_1} \frac{ \frac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z+ \sqrt{\frac{1}{3}} i)}}{(z-\sqrt{\frac{1}{3}}i)} dz + \displaystyle\oint_{C_2} \frac{ \frac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z- \sqrt{\frac{1}{3}} i)}}{(z+\sqrt{\frac{1}{3}}i)} dz \right)

=- \frac{4}{3}i \left(2\pi i \left[\dfrac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z+ \sqrt{\frac{1}{3}} i)}\right] _{z=\sqrt{\frac{1}{3}}i} + 2\pi i \left[\dfrac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z- \sqrt{\frac{1}{3}} i)}\right] _{z=-\sqrt{\frac{1}{3}}i}\right)

And tidy up the surds from there.

I've gotta say, this isn't the best integral to use this method on...

Reply 3

Original post by DFranklin

Where have the 'z's in the numerator of your last line of maths come from? (I may well just be being stupid and/or need to do some actual calculations, but I can't see it).

When the OP subbed in

\cos (t)

in terms of z, this produced a

\dfrac{1}{z}

term in the denominator.

Reply 4

DFranklin

Original post by Farhan.Hanif93

When the OP subbed in

\cos (t)

in terms of z, this produced a

\dfrac{1}{z}

term in the denominator.

I was talking about the change from the line just before

"considering singularities..."

and the line after.

(Looking a little closer, I think there may be a z missing from the numerator of the line before, but I'm not 100% sure).

Reply 5

Original post by DFranklin

I was talking about the change from the line just before

"considering singularities..."

and the line after.

(Looking a little closer, I think there may be a z missing from the numerator of the line before, but I'm not 100% sure).

Oh I see, yeah there should be a

z

in the numerator of the line before.

(edited 12 years ago)

Reply 6

Perpetuallity

OP

Original post by DFranklin

I was talking about the change from the line just before

"considering singularities..."

and the line after.

(Looking a little closer, I think there may be a z missing from the numerator of the line before, but I'm not 100% sure).

Yeah that's a typo, sorry, it should be a z.

Reply 7

Perpetuallity

OP

Original post by Farhan.Hanif93

I don't understand where you're stuck. It should be fairly obvious how to proceed from there. How well do you know Cauchy's integral formula? Notice that your poles are both of first order and since the singularities are at

z=\sqrt{\dfrac{1}{3}}i

and

z=-\sqrt{\dfrac{1}{3}}i

respectively for each term.

It follows that:

- \frac{4}{3}i\left( \displaystyle\oint_{C_1} \frac{ \frac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z+ \sqrt{\frac{1}{3}} i)}}{(z-\sqrt{\frac{1}{3}}i)} dz + \displaystyle\oint_{C_2} \frac{ \frac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z- \sqrt{\frac{1}{3}} i)}}{(z+\sqrt{\frac{1}{3}}i)} dz \right)

=- \frac{4}{3}i \left(2\pi i \left[\dfrac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z+ \sqrt{\frac{1}{3}} i)}\right] _{z=\sqrt{\frac{1}{3}}i} + 2\pi i \left[\dfrac{z}{(z + \sqrt{3}i)(z- \sqrt{3} i)(z- \sqrt{\frac{1}{3}} i)}\right] _{z=-\sqrt{\frac{1}{3}}i}\right)

And tidy up the surds from there.

I've gotta say, this isn't the best integral to use this method on...

Ah cheers. I see, now. The cauchy formula on my whiteboard had an error in it :facepalm2:

. I agree, It's a nasty integral for contour integration. Thanks!

Reply 8

nuodai

17

Original post by Farhan.Hanif93

...

Wow, you're already better at complex analysis than I am!

Reply 9

Original post by nuodai

Wow, you're already better at complex analysis than I am!

I wouldn't say I understand it too well! :wink:

If you're looking for something impressive, the OP is 16. :p:

Reply 10

Dirac Spinor

Have you shown that the contour integral has the same value as the real valued one? I'm not sure whether it is the case with circles in the complex plane, semicircles are much nicer.

Reply 11

DFranklin

Original post by ben-smith

Have you shown that the contour integral has the same value as the real valued one? I'm not sure whether it is the case with circles in the complex plane, semicircles are much nicer.

Barring any mistakes, yes. The integrand has been constructed so that going around the circle is "the same" as integrating the original integral from 0 to 2pi. (That's where the

\dfrac{dz}{iz}

has come from - the 'iz' is to cancel out that

dz

is

ie^{it} dt

).

Reply 12

Perpetuallity

OP

Original post by ben-smith

Have you shown that the contour integral has the same value as the real valued one? I'm not sure whether it is the case with circles in the complex plane, semicircles are much nicer.

I'm not convinced that a semicircular path would be 'nicer' in this particular problem. It may be due to the periodicity of the function, but I could be wrong...? :dontknow:

Reply 13

Dirac Spinor