Cauchy Integral

can you write z^2 + 9 as something of the form (z-z0) ?

 

You must specify what the path of integration is. There are singularities at z = 0, and z = ±3i.

The answer you gave is correct if the contour encloses only z = 0, and is traversed in the counter clockwise direction.

To find the residue at z = 0, you multiply the integrand by (z - 0) = z and evaluate the result at z = 0. If you do this, you will get 1/9.

 

hint: Yes you can do it, read dx's post - you also need to specify what the integration contour is

 

First of all, what is z₀? And why are you trying to get it into that form?

 

the strategy is to find all points in the complex plane where your integrand is singular, then you find out which one are lying inside your contour of integration.

Can you please write down the entire Cauchy's integral Theorem for us so we see that you are trying.

 

you have still not said anything what z0 is ... is it something special about it?

 

so which possible singularities do your integrand posses (hint: there are three such) and which one is closed within your contour??

 

One does not need to know the Residue theorem to evaluate this for single poles, in many textbooks Cauchy's formula is introduced earlier than the more general Residue theorem...

 

Oh, right. Deleted my prev. post.

 

As hinted previously, to find all singularities write [itex]z^2+9=(z-a)(z+a)[/itex]. The value a should be easy to find.

 

oh my, then you are really in trouble, i don't even know how to explain such simple thing xD

if you have cos(z)/[ z(z^2 + 9) ] and you want the the form f(z)/(z-0), can you GUESS??

 

There's already a z in the denominator. Just replace it with (z - 0).