MHB How to use cauchy integral formula

[aruwin]

Hello.
How do I know when to use Cauchy integral formula. Why do we use the formula in this question? As you can see in my attempt, I got stuck.
What is f(z), z, z_​0 here?

 

[chisigma]

Hello.
How do I know when to use Cauchy integral formula. Why do we use the formula in this question? As you can see in my attempt, I got stuck.
What is f(z), z, z_​0 here?

Instead of the Cauchy integral formula it is better for You to use the residue theorem that extablishes that...

$\displaystyle \int_{\gamma} f(z)\ dz = 2\ \pi\ i\ \sum_{j} r_{j}\ (1)$

... where $r_{j}$ is a residue of the j-th pole of f(*) inside $\gamma$. In your case the only pole inside $\gamma$ is z=-i and is...

$\displaystyle r_{1} = \lim_{z \rightarrow -i} \frac{(z+i)}{z^{2}+1} = \lim_{z \rightarrow -i} \frac{1}{z-i} = -\frac{1}{2\ i}\ (2)$

... so that is...

$\displaystyle \int_{\gamma} \frac{dz}{z^{2}+1}= - \pi\ (3)$

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Instead of the Cauchy integral formula it is better for You to use the residue theorem that extablishes that...

$\displaystyle \int_{\gamma} f(z)\ dz = 2\ \pi\ i\ \sum_{j} r_{j}\ (1)$

... where $r_{j}$ is a residue of the j-th pole of f(*) inside $\gamma$. In your case the only pole inside $\gamma$ is z=-i and is...

$\displaystyle r_{1} = \lim_{z \rightarrow -i} \frac{(z+i)}{z^{2}+1} = \lim_{z \rightarrow -i} \frac{1}{z-i} = -\frac{1}{2\ i}\ (2)$

... so that is...

$\displaystyle \int_{\gamma} \frac{dz}{z^{2}+1}= - \pi\ (3)$

Kind regards

$\chi$ $\sigma$

But I haven't learned about residue yet. So, I think I have to solve it using cauchy.

 

[chisigma]

But I haven't learned about residue yet. So, I think I have to solve it using cauchy.

Very well!... so we remember that the Cauchy integral formula supplies the value of an f(z) if z is inside a closed contour $\gamma$ in which f(z) is analitic...

$\displaystyle f(a) = \frac{1}{2\ \pi\ i} \int_{\gamma} \frac{f(z)}{z-a}\ d z\ (1)$

If You choose $\displaystyle f(z) = \frac{1}{z-i}$, $a=-i$ and apply (1) You obtain...

$\displaystyle \int_{\gamma} \frac{d z}{1 + z^{2}} = - 2\ \pi\ i\ \frac{1}{2\ i} = - \pi$

Kind regards

$\chi$ $\sigma$

 

[aruwin]

Very well!... so we remember that the Cauchy integral formula give the value of an f(z) if z is inside a closed contour $\gamma$ in which f(z) is analitic...

$\displaystyle f(a) = \frac{1}{2\ \pi\ i} \int_{\gamma} \frac{f(z)}{z-a}\ d z\ (1)$

If You choose $\displaystyle f(z) = \frac{1}{z-i}$, $a=-i$ and apply (1) You obtain...

$\displaystyle \int_{\gamma} \frac{d z}{1 + z^{2}} = - 2\ \pi\ i\ \frac{1}{2\ i} = - \pi$

Kind regards

$\chi$ $\sigma$

I got it! Thanks!