Complicated integration of complex number

[MissP.25_5]

Hello.
I am not confident about this question. I think I have to use cauchy integral formula. But before that, I should decompose the fraction, right? Or is there a simpler way to do it?　A friend told me that each contour only had one pole interior to it so he just used the Cauchy integral formula for the appropriate derivatives--as opposed to using a residue calculus. Is this a correct method?

 

[CAF123]

You can first sketch the contour and identify the nature and location of the isolated singularities of f(z). Then the evaluation of the integral is immediate by using the residue theorem.

Yes, you can use the Cauchy integral formula too.

 

[MissP.25_5]

You can first sketch the contour and identify the nature and location of the isolated singularities of f(z). Then the evaluation of the integral is immediate by using the residue theorem.

Yes, you can use the Cauchy integral formula too.

As for the first one, z=2i is outside of the circle, since the circle is a unit circle. That means f(z) is analytic inside the circle, right? Hence we can use Cauchy integral formula.
But for the second one, z=2i is inside the circle, we can't use Cauchy integral formula, can we?

 

[CAF123]

As for the first one, z=2i is outside of the circle, since the circle is a unit circle. That means f(z) is analytic inside the circle, right?

No, because there is also a pole at the origin, which lies inside the unit circle.

But for the second one, z=2i is inside the circle, we can't use Cauchy integral formula, can we?

There is Cauchy's integral formula but also Cauchy's integral theorem. CIT is just a [STRIKE]generalisation[/STRIKE] special case of the residue theorem.

 

[MissP.25_5]

No, because there is also a pole at the origin, which lies inside the unit circle.


There is Cauchy's integral formula but also Cauchy's integral theorem. CIT is just a generalisation of the residue theorem.

So does that mean CIF can be used even there's a pole inside a circle? And I am not sure if I did this right, can you check? I did it a little so you can check if it's correct so far.

 

[CAF123]

So does that mean CIF can be used even there's a pole inside a circle? And I am not sure if I did this right, can you check? I did it a little so you can check if it's correct so far.

Suppose your integral is $\oint_C f(z) dz$. You need to pick a function g(z) that is analytic in the interior of the unit circle. Then, loosely speaking, the part of f(z) that is not analytic in the unit circle picks up a residue when you integrate over the boundary. To illustrate: you have $$\oint \frac{e^{z^2}}{z^3 (z-2i)^2} dz = \oint \frac{g(z)}{z^3}dz$$ g(z) is analytic in the interior of the unit circle. 1/z³ is not, so you pick up a residue, which is what appears on the other side of CIF. If f(z) happened to be completely analytic in the interior, then there is no residue, so the other side of CIF collapses to zero, I.e recovering CIT.

The residue theorem is a complete generalisation of both CIT and CIF so you can always use that.

 

[MissP.25_5]

Suppose your integral is $\oint_C f(z) dz$. You need to pick a function g(z) that is analytic in the interior of the unit circle. Then, loosely speaking, the part of f(z) that is not analytic in the unit circle picks up a residue when you integrate over the boundary. To illustrate: you have $$\oint \frac{e^{z^2}}{z^3 (z-2i)^2} dz = \oint \frac{g(z)}{z^3}dz$$ g(z) is analytic in the interior of the unit circle. 1/z³ is not, so you pick up a residue, which is what appears on the other side of CIF. If f(z) happened to be completely analytic in the interior, then there is no residue, so the other side of CIF collapses to zero, I.e recovering CIT.

The residue theorem is a complete generalisation of both CIT and CIF so you can always use that.

Ok, so this is what I got so far and now how do I continue? I don't know how to apply CIF after decomposing the fraction.

 

[CAF123]

You did not have to split the fraction up. But to proceed, write your last step as a sum of three integrals. In each one, identify g(z) that is analytic in the interior of the unit circle and apply CIF. (g(z) is the f(z) that appears in CIF, but you already used the notation f(z) to mean the function you are integrating over the boundary.)

 

[MissP.25_5]

You did not have to split the fraction up. But to proceed, write your last step as a sum of three integrals. In each one, identify g(z) that is analytic in the interior of the unit circle and apply CIF. (g(z) is the f(z) that appears in CIF, but you already used the notation f(z) to mean the function you are integrating over the boundary.)

I didn't have to decompose the fraction? Then what is the easiest way? Please tell me.

 

[MissP.25_5]

Suppose your integral is $\oint_C f(z) dz$. You need to pick a function g(z) that is analytic in the interior of the unit circle. Then, loosely speaking, the part of f(z) that is not analytic in the unit circle picks up a residue when you integrate over the boundary. To illustrate: you have $$\oint \frac{e^{z^2}}{z^3 (z-2i)^2} dz = \oint \frac{g(z)}{z^3}dz$$ g(z) is analytic in the interior of the unit circle. 1/z³ is not, so you pick up a residue, which is what appears on the other side of CIF. If f(z) happened to be completely analytic in the interior, then there is no residue, so the other side of CIF collapses to zero, I.e recovering CIT.

The residue theorem is a complete generalisation of both CIT and CIF so you can always use that.

I'm trying to use what you said here, but I don't really understand how to do this. In this case, g(z) should be
e^(z²)/(z - 2i)², right? g(z) is actually f(z), isn't it?

 

[CAF123]

In this case, g(z) should be
e^(z²)/(z - 2i)², right?

Right.

g(z) is actually f(z), isn't it?

If you write CIF as $$f^n(z_o) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_o)^{n+1}}dz$$ then yes, this choice of g(z) coincides with that f(z) in CIF.

 

[MissP.25_5]

Right.
If you write CIF as $$f^n(z_o) = \frac{n!}{2\pi i} \oint \frac{f(z)}{(z-z_o)^{n+1}}dz$$ then yes, this choice of g(z) coincides with that f(z) in CIF.

n here is 2, right? Because the non-analytic part is 1/z^3

 

[CAF123]

n here is 2, right? Because the non-analytic part is 1/z^3

Correct and $z_o = 0$.

 

[MissP.25_5]

Correct and $z_o = 0$.

I am beginning to understand it now. So, I have to differentiate f(z) twice and multiply it with ∏i?
The derivative part is kind of tedious.

 

[CAF123]

I am beginning to understand it now. So, I have to differentiate f(z) twice and multiply it with ∏i?
The derivative part is kind of tedious.

If you mean the f(z) that appears in CIF then yes. The derivatives are tedious as you say, especially for higher order poles.

The same result can be obtained using the residue theorem: $$\oint_C f(z) dz = 2 \pi i \sum_{z_k \in \text{Int} C} \text{Res}(f, z_k)$$

In this case, $z = 0$ is a pole of order 3 and is the only singularity located in the interior of the unit circle.
So the RHS of the residue theorem is $$2 \pi i \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d}{dz} z^3 f(z)$$ Put in f(z), simplify and take derivative. (here f(z) is $e^{z^2}/z^3 (z-2i)^2$)

 

[MissP.25_5]

If you mean the f(z) that appears in CIF then yes. The derivatives are tedious as you say, especially for higher order poles.

The same result can be obtained using the residue theorem: $$\oint_C f(z) dz = 2 \pi i \sum_{z_k \in C} \text{Res}(f, z_k)$$

In this case, $z = 0$ is a pole of order 3 and is the only singularity located in the interior of the unit circle.
So the RHS of the residue theorem is $$2 \pi i \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d}{dz} z^3 f(z)$$ Put in f(z), simplify and take derivative. (here f(z) is $e^{z^2}/z^3 (z-2i)^2$)

I don't understand this residue part. If I use the above formula, then there is no need to use CIF??

 

[CAF123]

I don't understand this residue part. If I use the above formula, then there is no need to use CIF??

Exactly, CIF is 'contained' within the Residue theorem and CIT is a special case. So really if you know the residue theorem you are good to go with all these types of problems.

Depending on the order of the pole, the form for the residue will change. They are derived using the Laurent series of the function. See your textbook for derivation and general formula.

 

[MissP.25_5]

Exactly, CIF is 'contained' within the Residue theorem and CIT is a special case. So really if you know the residue theorem you are good to go with all these types of problems.

Depending on the order of the pole, the form for the residue will change. They are derived using the Laurent series of the function. See your textbook for derivation and general formula.

I have been differentiating this but I don't think this is right. Can you tell me where i am wrong?

 

[CAF123]

I have been differentiating this but I don't think this is right. Can you tell me where i am wrong?

The first derivative looks right, but I think you skipped a few steps in the second derivative so it is difficult for me to see where you went wrong. In any case, provided you understand the method that is more or less the answer you are looking for

What is the numerical answer provided anyway?

 

[MissP.25_5]

The first derivative looks right, but I think you skipped a few steps in the second derivative so it is difficult for me to see where you went wrong. In any case, provided you understand the method that is more or less the answer you are looking for

What is the numerical answer provided anyway?

-∏i/8

 

[MissP.25_5]

The first derivative looks right, but I think you skipped a few steps in the second derivative so it is difficult for me to see where you went wrong. In any case, provided you understand the method that is more or less the answer you are looking for

What is the numerical answer provided anyway?

By the way, could you show me how to use the residue theorem to solve this?

 

[CAF123]

-∏i/8

I confirmed this result. I suggest redoing or checking your work again from the point where you start to take the second derivative. Most likely small errors. Also 2ze^z2 should be regarded as two functions of z.

By the way, could you show me how to use the residue theorem to solve this?

As you can see from post #15, the expression you deal with is exactly the same as that you are dealing with right now.

 

[MissP.25_5]

I confirmed this result. I suggest redoing or checking your work again from the point where you start to take the second derivative. Most likely small errors. Also 2ze^z2 should be regarded as two functions of z.As you can see from post #15, the expression you deal with is exactly the same as that you are dealing with right now.

That's for no. 1, right?

 

[MissP.25_5]

If you mean the f(z) that appears in CIF then yes. The derivatives are tedious as you say, especially for higher order poles.

The same result can be obtained using the residue theorem: $$\oint_C f(z) dz = 2 \pi i \sum_{z_k \in \text{Int} C} \text{Res}(f, z_k)$$

In this case, $z = 0$ is a pole of order 3 and is the only singularity located in the interior of the unit circle.
So the RHS of the residue theorem is $$2 \pi i \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d}{dz} z^3 f(z)$$ Put in f(z), simplify and take derivative. (here f(z) is $e^{z^2}/z^3 (z-2i)^2$)

Take derivative of what?

 

[CAF123]

That's for no. 1, right?

no. 2 is done in exactly the same way, except now z=0 is not in the interior of the contour, but z = 2i is. And this is a pole of order 2, so in the end you compute one less derivative.

 

[CAF123]

Take derivative of what?

(..Typo in previous post, should have two derivatives..)

The RHS of the residue theorem is $$2\pi i \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{dz^2} z^3 f(z) = \pi i \lim_{z \rightarrow 0} \frac{d^2}{dz^2} z^3 \frac{e^{z^2}}{z^3 (z-2i)^2} = \pi i \lim_{z \rightarrow 0} \frac{d^2}{dz^2} \frac{e^{z^2}}{(z-2i)^2}$$

Both numerator and denominator are continuous functions of z at z=0, (as is expected at this stage) so simply plug in z=0 after computing two derivatives. Exactly what you just did with CIF.

 

[MissP.25_5]

no. 2 is done in exactly the same way, except now z=0 is not in the interior of the contour, but z = 2i is. And this is a pole of order 2, so in the end you compute one less derivative.

Is this ok?

 

[CAF123]

Is this ok?

See above, there was a typo, sorry!

 

[MissP.25_5]

(..Typo in previous post, should have two derivatives..)

The RHS of the residue theorem is $$2\pi i \frac{1}{2!} \lim_{z \rightarrow 0} \frac{d^2}{dz^2} z^3 f(z) = \pi i \lim_{z \rightarrow 0} \frac{d^2}{dz^2} z^3 \frac{e^{z^2}}{z^3 (z-2i)^2} = \pi i \lim_{z \rightarrow 0} \frac{d^2}{dz^2} \frac{e^{z^2}}{(z-2i)^2}$$

Both numerator and denominator are continuous functions of z at z=0, (as is expected at this stage) so simply plug in z=0 after computing two derivatives. Exactly what you just did with CIF.

You mean I have to differentiate that twice just like I did with CIF? Then it will be tedious also?

 

[CAF123]

You mean I have to differentiate that twice just like I did with CIF? Then it will be tedious also?

Yes, but you already done the computation using CIF so need to bother doing it again.

 

[MissP.25_5]

Yes, but you already done the computation using CIF so need to bother doing it again.

The it means the this wasy is no simpler than CIF? Because I still haven't finished the derivation.

 

[CAF123]

The it means the this wasy is no simpler than CIF? Because I still haven't finished the derivation.

Well, it is a matter of taste but in some sense I find the residue theorem easier to actually get to that equation involving the derivative. You did not have to, for example, study f(z) to find a part that is analytic on the interior of the contour. Even though that is still quite a quick step...

Laurent expansions of functions also give you the residue - it is defined as the coefficient of the 1/z term (the first term in the expansion that 'knows' about the singularity). But again, you end up with the final equation and in fact this is how the general formula for the residue of a nth order pole was derived. If you have a textbook or notes, check it out.