How Do You Solve Complex Integral and Laurent Series Problems?

[Mappe]

Need help with complex problems!

I have trouble with a residue problem:

Integral -inf --> inf sin(ax)/(x^2 + 4x + 5)

As both it's poles contains both a real and a complex side, the sin(ax) part gets very ugly when I try to calculate it's residues. And I get a complex answere! Do I just ignore the complex part of the answere, and why?

Also, I can't figure out how to make a laurentseries out of:

1/(1-cos(x))

Thanx for any help!

 

[Dick]

No, you don't just ignore the complex parts. If you did it right, they should cancel on their own. You'll need to split sin(ax) up into (exp(iax)-exp(-iax))/(2i) since you have to close the contour in different half planes for each of those parts depending on the sign of a. Show some more of your work!

 

[Mappe]

Well, so I choose the residue at (-2 + i), and thereby the first and second quadrant.
In the (z +2 -i)(z +2 +i) term, the first one gets canceled, and inserting my chosen number in the second gives 1/(2i) (cuz its in the denominator). This is canceled out when I rewrite the sin part. Now I have: (e^(az)-e^(-az))*(pi)/(2i) (switch down the i to cancel the minus)

As my selected satisfies |a|>0 and |b|>0 in a+ib, the (e^(az) - e^(-az)) has a complex component and a real one. Doesnt that leaves me with a complex answer?

 

[Dick]

Well, so I choose the residue at (-2 + i), and thereby the first and second quadrant.
In the (z +2 -i)(z +2 +i) term, the first one gets canceled, and inserting my chosen number in the second gives 1/(2i) (cuz its in the denominator). This is canceled out when I rewrite the sin part. Now I have: (e^(az)-e^(-az))*(pi)/(2i) (switch down the i to cancel the minus)

As my selected satisfies |a|>0 and |b|>0 in a+ib, the (e^(az) - e^(-az)) has a complex component and a real one. Doesnt that leaves me with a complex answer?

You are going way too fast. Like I said the exp(iaz) part and the exp(-iaz) parts need contours in different half planes. You need to split it into two contour integrals and do each one separately, then add them to get the final answer. The complex parts will only disappear when you add them together.

 

[Mappe]

Ok, thanx. How come I can use the exp function in this way, but not the sin function? in other words, why do I need to expand the sin function but not the exp function to get the residue? And also, why does the contour integrals have to be in different half planes?

 

[Dick]

Ok, thanx. How come I can use the exp function in this way, but not the sin function? in other words, why do I need to expand the sin function but not the exp function to get the residue? And also, why does the contour integrals have to be in different half planes?

If you take a>0 then exp(iaz) goes to zero exponentially as the imaginary part of z goes to infinity in the UPPER half plane. Look at the real part of exp(ia(c+di)) as d->+infinity. So you complete that contour in the upper half you can ignore the contribution from the semicircle. Similary exp(-iaz) is well behaved in the LOWER half plane. sin(az) is badly behaved in BOTH half planes. That's why you need to split it.

 

[Mappe]

Ooh, very nice. Trying it out.

 

[Mappe]

Ok, I've tried it now, and I still get it imaginary.

sin(az)/(z^2 + 4z + 5) = sin(az)/((z+2-i)(z+2+i)) =
(1/2i)*(e^(iaz)/ ((z+2-i)(z+2+i))) - (1/2i)*(e^(-iaz)/ ((z+2-i)(z+2+i)))

And now choosing poles, the e^(iaz) needs to have the upper half-plane, and the (-2+i) residue, and the other pole for e^(-iaz). This all gives:

(1/2i)*(e^(-a - 2ai))*(1/2i) - (1/2i)*(e^(-a + 2ai))*(1/-2i)

Cancel the middle-minus with the one in the last denominator, and cancel both 1/i terms:

-(1/4)(e^(-a - 2ai) + e^(-a + 2ai))

In there, we see that the complex parts of both exp functions are the same, but different signs, and the real is the same completely. This means that the sin, and there by the imaginary, part of the definition of exp cancels. Left we have:

-(1/4)*e^(-a)*2cos(2a)

Then we multiply that with 2(pi)i, and the result is complex. Please help!

 

[Dick]

I don't have time to check the details right now, that seems roughly ok. But look back at your contours. The one in the upper half plane has to run counterclockwise and the one in the lower half plane has to be clockwise. When you apply the residue theorem you have to change a sign to compensate for that.

 

[Mappe]

YES! thanx. How come it has to be clockwise?

 

[Dick]

Both contours have to go in the positive direction along the real axis (since the integral is from -infinity to +infinity. Then to complete the contour in the upper half plane your big arc has to be counterclockwise. Just the opposite in the lower half plane.