Thread: Strange complex integral/path question

1. Strange complex integral/path question

Hey guys,

I've been trying to work out what this textbook question means! Well the first part that is, is it supposed to be say we must find those constants before proceeding? I've tried adding them all together just alebratically but its ugly and also integrating each fraction over the path but I feel like these constant need to vanish. It's an odd number so there are no solutions provided. I thought I would be able to use the same logic as the previous question, but its completely different! I'm really struggling with this one, enough to have made an account on here and ask for help!
Could anyone point me in the right direct?

Thanks alot,

2. Re: Strange complex integral/path question

Originally Posted by Angela11
I interpret that you have to express the integral in terms of the given constants. Taking into account that $\displaystyle \pi<\sqrt{10}<2\pi$, we have

$\displaystyle (i)\;\int_{|z|=\pi}\frac{dz}{z^3(z^2+10)}=P\int_{| z|=\pi}\frac{dz}{z}+Q\int_{|z|=\pi}\frac{dz}{z^2}+ R\int_{|z|=\pi}\frac{dz}{z^3}=\ldots$

$\displaystyle (ii)\;\int_{|z|=2\pi}\frac{dz}{z^3(z^2+10)}=P\int_ {|z|=2\pi}\frac{dz}{z}+\ldots +T\int_{|z|=2\pi}\frac{dz}{z+\sqrt{10}i}=\ldots$

Now, apply the Cauchy integral formula.

3. Re: Strange complex integral/path question

Thanks,

I did something similar to that though I let z = pie^(it), so dz/dt = i pi e^(it) and for the P integral i got 2ipiP, and so on, then I got stuck at the S and T parts, I wasn't sure if f(z) was just = S or T.
If f(z)=S, say, would f(-isqrt10) just = S, how come the S and T parts cancelled in part i? I tried rationalising and adding the S and T parts together but it didn't look helpful

4. Re: Strange complex integral/path question

Originally Posted by Angela11
and for the P integral i got 2ipiP
Right, but before continuing: have you covered the Cauchy integral formula?.

5. Re: Strange complex integral/path question

Briefly,

I know that int over c is f(z)/(z-z0) = 2ipif(z0) if z0 is enclosed within C, But we have more been using that approach above, kind of parametrizing first.

We have used Cauchys integral formula as far as find the integral of 2z/(z-3) for example,

So i thought for S the z0 would be isqrt(10) so the integral would be -2pisqrt(10) or S2ipi depending on if f(z)=S and if f(isqrt(10)) just = S,

Just a little confused at this point

6. Re: Strange complex integral/path question

Well, using the Cauchy integral formula or parametrizing, you'll easily obtain:

$\displaystyle \int_{|z|=\pi}\frac{dz}{z^3(z^2+10)}=2P\pi i$

$\displaystyle \int_{|z|=2\pi}\frac{dz}{z^3(z^2+10)}=2P\pi i+2S\pi i\color{red}+\color{black}2T\pi i$

7. Re: Strange complex integral/path question

I understand that the Q and R parts are zero, but i'm confused with the S and T parts,

for ii)

Is this the correct logic for doing the S and T parts, letting the Cauchy integral formula be int f(z)/(z-a) = 2ipif(a)

S:
a = isqrt(10), f(z) = S, f(a) = S so the integral is 2Sipi

T:

a=-isqurt(10) f(z)=T, f(-sqrt(10))=-f(sqrt(10))=-T so -2Tipi

and for part i, are the S and T integrals zero as pi<sqrt(10) so a=isqrt(10) is outside |z|=pi

8. Re: Strange complex integral/path question

Originally Posted by Angela11
for ii) Is this the correct logic for doing the S and T parts, letting the Cauchy integral formula be int f(z)/(z-a) = 2ipif(a)
Yes.

S: a = isqrt(10), f(z) = S, f(a) = S so the integral is 2Sipi
Right.

T: a=-isqurt(10) f(z)=T, f(-sqrt(10))=-f(sqrt(10))=-T so -2Tipi
No, it is $\displaystyle f(a)=T$ (I had a typo in my previous post)

and for part i, are the S and T integrals zero as pi<sqrt(10) so a=isqrt(10) is outside |z|=pi
Right.

9. Re: Strange complex integral/path question

Many thanks Fernando!