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I am having trouble answering the following question:

Problem

By integrating the function:

$$f(z) = \frac{R + z}{z(R - z)}$$

round a suitable contour, prove that, for $0 \leq r < R$,

$$\frac{1}{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta = 1 $$


My Attempt

What I think they want me to do is to use the equation:

$$\int_C f(z) \mathrm{d}z = \int_{z_1}^{z_2} f(z) \mathrm{d}z = \int_a^bf(z(t))\cdot z'(t)\mathrm{d}t $$

where C is a contour from $z_1$ to $z_2$ and $z(a) = z_1$ and $z(b) = z_2$. Then I should find a contour $z(t)$ in terms of $r$ and $t$ such that:

$$f(z(t))\cdot z'(t) = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2} $$

And then with this contour I could find values for $z_1$ and $z_2$ meaning I could evaluate:

$$ \int_{z_1}^{z_2} f(z) \mathrm{d}z $$

which should equal 2$\pi$ or something. However so far I have not been able to find a suitable z(t). The closest I have come is $z(t) = re^{i\theta}$ which yields:

$$f(z(t))\cdot z'(t) = \frac{R^2\cdot i - 2Rr\sin(t) + r - r^2\cdot i}{R^2 - 2Rr\cos(t) + r^2} $$

However this entire approach could be wrong. I've got to the stage when you start to question your whole approach and feel like resorting to violence...According to Einstein, only insane people do the same thing over and over again and expect different results......Oh wait that's me...

Please help me.


Edit 1


First of all a big thank you to everyone who has helped me here - your hints are like a life raft to a drowning man!

To Mariano: I have not done anything on residues at this stage in the course - so the question should be answerable without using them. That technique does look very powerful though and I hope I will get a good understanding of it in the future...

To Hans Lundmark: Your link helped me to see the bigger picture put this equation in context (wow its more that a figment of my prof's imagination!)...Thanks.

To J. M: Your note helped a lot to remove the log jam in my head but there are still some issues that remain with my answer. I would very much appreciate your comments on my 2nd attempt.

2nd Attempt

(Step 1)

The function $f(z)$ can be decomposed using partial fractions:

$$f(z) = \frac{R + z}{z(R - z)} = \frac{1}{z} + \frac{2}{R - z}$$

and then integrated with respect to $z$:

$$\int f(z)dz = log(z) - 2log(R - z) $$

(Step 2)

With this information we can evaluate:

$$\int_C f(z)dz $$

where C is the positively oriented circle $z = re^{i\theta}$. Because log(z) is not defined at its branch cut we cannot integrate around the whole circle but must split the circle into halves $C_1$ and $C_2$.

Let $C_1$ denote the right half of the circle where $ -\frac{\pi}{2} <= \theta <= \frac{\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $-\pi <= \theta <= \pi$. Therefore:

$$ \int_{C_1} f(z) = [log(z) - 2log(R - z)]_{-ri}^{ri} $$

which equals:

$$ \int_{C_1} f(z)dz = \pi{i} - 2log(R - ri) + 2log(R + ri) $$

Let $C_2$ denote the left half of the circle where $\frac{\pi}{2} <= \theta <= \frac{3\pi}{2}$ and consider the branch: $log(z) = ln(r) + i\theta$ where $0 <= \theta <= 2\pi$. Therefore:

$$ \int_{C_2} f(z) = [log(z) - 2log(R - z)]_{ri}^{-ri} $$

which equals:

$$ \int_{C_2} f(z)dz = \pi{i} - 2log(R + ri) + 2log(R - ri) $$

and since:

$$ \int_{C} f(z)dz = \int_{C_1} f(z)dz + \int_{C_2} f(z)dz $$

it follows that:

$$ \int_{C} f(z)dz = 2\pi{i} $$

(Step 3)

This is the part I am currently having difficulty with:

Does the equation:

$$\int_C f(z) \mathrm{d}z = \int_a^bf(z(t))\cdot f'(t) dt $$

imply that:

$$\Im\int_C f(z) \mathrm{d}z = \int_a^b\Im[{f(z(t))}\cdot f'(t) dt] $$

Because if it did I could say that:

$$ \Im(2\pi{i}) = \int_0^{2\pi} \frac{R^2-r^2}{R^2 - 2Rr\cos(\theta) + r^2}\mathrm{d}\theta $$

and the proof would be complete...But I dont know if I can do this?

Can some one help please?


Edit 2


Thanks to J.M step 3 should be:

Since the integral operator is linear:

$$ \int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$

therfore:

$$ \int_C f(z) \mathrm{d}z = \int_a^b\Re{f(z(t))}\cdot f'(t) dt + i\int_a^b\Im{f(z(t))}\cdot f'(t) dt $$

also we know that:

$$ \int_{0}^{2\pi}\Re[{f(z(t))}\cdot f'(t) dt] = 0 $$

and:

$$ \int_C f(z) \mathrm{d}z = 2{\pi}i $$

and:

$$ \Im[f(z(t))\cdot z'(t)] = \frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2} $$

Therefore:

$$ \int_C f(z) \mathrm{d}z = i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$

$$ 2{\pi}i = i\int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$

$$ 2{\pi} = \int_0^{2\pi}\Im[{f(z(t))}\cdot f'(t)] dt $$

$$ 2{\pi} = \int_0^{2\pi}\frac{R^2-r^2}{R^2 - 2Rr\cos(t) + r^2}dt $$

and the proof is complete. Awesome!

I hope this helps some other dude out there who is trying to understand this stuff...

share|improve this question
    
Do you know about residues? –  Mariano Suárez-Alvarez Sep 23 '10 at 8:49
    
You're on the right track, compare my answer with your last expression; then use Mariano's hint. –  J. M. Sep 23 '10 at 9:04
1  
en.wikipedia.org/wiki/Poisson_kernel#On_the_unit_disc might be of some interest too. If you integrate termwise in the series given there, only the term for $n=0$ contributes. (But of course this is not the method that you were asked to use here.) –  Hans Lundmark Sep 23 '10 at 10:48
1  
Now, this is the sort of beginner questions I like to see! Being proactive instead of being a sitting duck is a virtue! –  J. M. Sep 24 '10 at 7:44
    
-1 for the second sentence in the title. Psychatric illnesses are as serious as physical ones. Trivialising them seems insensitive to me. And then you talk about resorting to violence... All for the solution to one exercise! –  Jyotirmoy Bhattacharya Sep 24 '10 at 9:32

1 Answer 1

up vote 3 down vote accepted

Note that

$$\Im\left(\frac{R+r\exp(i\theta)}{r\exp(i\theta)(R-r\exp(i\theta))}(i r\exp(i\theta))\right)=\frac{R^2-r^2}{R^2+r^2-2Rr\cos\theta}$$


As for your second attempt, remember that the integral is a linear operator, so

$$\int z(t) \mathrm{d}t=\int \Re z(t) \mathrm{d}t+i\int \Im z(t) \mathrm{d}t$$

share|improve this answer
    
After this, have a gander at this MathWorld entry. –  J. M. Sep 23 '10 at 9:10
    
Woooooooppie Ive finally got it! Thanks so much. I would never have been able to solve this without your help! –  Johann Sep 24 '10 at 8:14
1  
It would be good if you could check my new answer - just so I don't mislead someone else. Cheers –  Johann Sep 24 '10 at 8:37

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