MHB Complicated integration of complex number

[aruwin]

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?

 

[chisigma]

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?

Very well!... I suppose that also in this case the residue theorem is 'forbidden', so that You have to use the general Cauchy integral formula...

$\displaystyle f^{(n)} (a) = \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}\ dz\ (1)$

If You choose $\displaystyle f(z)= \frac{e^{z^{2}}}{(z-2\ i)^{2}}$ with a little of patience You find that is...

$\displaystyle \frac{d^{2}}{d z^{2}} f(z) = 2\ e^{z^{2}}\ \frac{- 8\ z^{2} - 4 - 8\ i\ z^{3} - 4\ i\ z + 32 z^{4} - 3\ z + 3}{(z-2\ i)^{4}} = \frac{1}{8}\text {for}\ z=0\ (2)$

... so that is...

$\displaystyle \int_{\gamma} \frac{e^{z^{2}}}{z^{3}\ (z - 2\ i)^{2}}\ dz = \frac{\pi\ i}{8}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Very well!... I suppose that also in this case the residue theorem is 'forbidden', so that You have to use the general Cauchy integral formula...

$\displaystyle f^{(n)} (a) = \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}\ dz\ (1)$

If You choose $\displaystyle f(z)= \frac{e^{z^{2}}}{(z-2\ i)^{2}}$ with a little of patience You find that is...

$\displaystyle \frac{d^{2}}{d z^{2}} f(z) = 2\ e^{z^{2}}\ \frac{- 8\ z^{2} - 4 - 8\ i\ z^{3} - 4\ i\ z + 32 z^{4} - 3\ z + 3}{(z-2\ i)^{4}} = \frac{1}{8}\text {for}\ z=0\ (2)$

... so that is...

$\displaystyle \int_{\gamma} \frac{e^{z^{2}}}{z^{3}\ (z - 2\ i)^{2}}\ dz = \frac{\pi\ i}{8}\ (3)$

Kind regards

$\chi$ $\sigma$

Your f(z) is different from mine. Is mine wrong?

 

[Prove It]

In the first one, since the pole at z = 0 (I'm assuming that the unit circle is centred at the origin) is of order 3, you evaluate its residue using $\displaystyle \begin{align*} \textrm{Res}\,\left( f, c \right) = \frac{1}{2!} \lim_{z\to c} \frac{\mathrm{d}^2}{\mathrm{d}z^2} \left[ (z - c)^3\,f(z) \right] \end{align*}$.

Then the integral is calculated as $\displaystyle \begin{align*} \oint_C{ f(z)\,\mathrm{d}z } = 2\pi i \, \textrm{Res}\, \left( f, c \right) \end{align*}$.

 

[chisigma]

Your f(z) is different from mine. Is mine wrong?

What I undestood is that Your question (1) is to compute...

$\displaystyle \int_{\gamma} \frac{e^{z^{2}}}{z^{3}\ (z-2\ i)^{2}}\ (1)$

... where $\gamma$ is the unit circle... isn't it?...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

What I undestood is that Your question (1) is to compute...

$\displaystyle \int_{\gamma} \frac{e^{z^{2}}}{z^{3}\ (z-2\ i)^{2}}\ (1)$

... where $\gamma$ is the unit circle... isn't it?...

Kind regards

$\chi$ $\sigma$

Yes, but then why f(z) = exp(z^2)/(z-2i)^2 ? I don't understand.

 

[Prove It]

You are making life difficult on yourself. The Residue Theorem is designed to make calculating these complex contour integrals easier. Use it!

 

[aruwin]

You are making life difficult on yourself. The Residue Theorem is designed to make calculating these complex contour integrals easier. Use it!

No, I haven't learned that in class yet. So I have to use CIF.

 

[chisigma]

Yes, but then why f(z) = exp(z^2)/(z-2i)^2 ? I don't understand.

I apologize because I've been in a hurry!... if You observe with care the general Cauchy formula I'm sure that it will be for You fully clear!...

$\displaystyle f^{n} (a) = \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}}\ dz\ (1)$

Now suppose $\displaystyle f(z)= \frac{e^{z^{2}}}{(z-2\ i)^{2}}$ and a=0...

Kind regards

$\chi$ $\sigma$

 

[aruwin]

You are making life difficult on yourself. The Residue Theorem is designed to make calculating these complex contour integrals easier. Use it!

Can you tell me how to use the residue theorem here?

 

[chisigma]

It is not difficult to realize that the use of Cauchy integral formula and the Residue theorem, if correctly applied, lead to the same results, so that they are absolutely equivalent...

Kind regards

$\chi$ $\sigma$

 

[Prove It]

Can you tell me how to use the residue theorem here?

Read my previous posts in this thread...