Residue of z/cos(z): Explanation & Calculation

[pivoxa15]

The definition of a residue is the coefficient of the -1 power in the Laruent series. If I do z/cos(z) by long division, I get a series starting with z so z^-1 never occurs hence has a coefficient of 0. But why does it have a non zero residue, namely z/sin(z) at each z when cos(z)=0?

 

[TD]

But why does it have a non zero residue, namely z/sin(z) at each z when cos(z)=0?

I don't understand what you mean with this, but at z = 0, z/cos(z) has a 0 residue indeed...

 

[pivoxa15]

The residue occur only when cos(z)=0 so no residue exist when z=0 as cos(0)=1.

 

[matt grime]

z/sin(z) is not a residue, it is a function.

 

[pivoxa15]

I should have said residue of z/cos(z) when z=n(pie)/2 where n is odd.

 

[ircdan]

In general, if f(z) = P(z)/Q(z) where P and Q are analytic at z_0 , P(z_0) != 0, and Q(z) has a simple zero at z_0, then Res(f; z_0) = P(z_0)/Q'(z_0). For you, f(z) = z/cosz and cosz = 0 iff z = k*pi + pi/2, k = 0, +/- 1, +/- 2, ..., so that Res(f; z_0 = k*pi + pi/2) = (k*pi + pi/2)/(-sin(k*pi + pi/2)) for k = 0, +/- 1, +/-2, ...

If you don't have this result, you can prove it easily. Note since f has a simple pole at z_0 and Q(z_0) = 0, Res(f; z_0) = lim z->z_0 (z - z_0)*P(z)/Q(z) = lim z->z_0 P(z)/((Q(z) - Q(z_0))/(z - z_0)) = P(z_0)/Q'(z_0).

 

[pivoxa15]

In general, if f(z) = P(z)/Q(z) where P and Q are analytic at z_0 , P(z_0) != 0, and Q(z) has a simple zero at z_0, then Res(f; z_0) = P(z_0)/Q'(z_0). For you, f(z) = z/cosz and cosz = 0 iff z = k*pi + pi/2, k = 0, +/- 1, +/- 2, ..., so that Res(f; z_0 = k*pi + pi/2) = (k*pi + pi/2)/(-sin(k*pi + pi/2)) for k = 0, +/- 1, +/-2, ...

If you don't have this result, you can prove it easily. Note since f has a simple pole at z_0 and Q(z_0) = 0, Res(f; z_0) = lim z->z_0 (z - z_0)*P(z)/Q(z) = lim z->z_0 P(z)/((Q(z) - Q(z_0))/(z - z_0)) = P(z_0)/Q'(z_0).

What about my long division and getting z as the first term in the series of z/cos(z)? Hence no z^-1 term?

 

[matt grime]

But are you using the correct expansion? The Laurent series at 0 tells you nothing about the residues at poles away from zero (which is where all the poles are)

 

[pivoxa15]

But are you using the correct expansion? The Laurent series at 0 tells you nothing about the residues at poles away from zero (which is where all the poles are)

You are correct but how do you know I expanded the Laurent series about z=0?

Even if I exand the Laurent series with centre at the first 0 which is pie/2, after the long division, I still get a series starting with z. hence no z^-1 term.

 

[matt grime]

You are correct but how do you know I expanded the Laurent series about z=0?

it was probable from the way you wrote your reply.

Even if I exand the Laurent series with centre at the first 0 which is pie/2, after the long division, I still get a series starting with z. hence no z^-1 term.

Then you have expanded it incorrectly: one object has a pole the other does not. (I'm presuming you have no other negative powers of z).

 

[pivoxa15]

it was probable from the way you wrote your reply.



Then you have expanded it incorrectly: one object has a pole the other does not. (I'm presuming you have no other negative powers of z).

Do you think it is feasible to do by hand? Since we have z/(a series involving terms like z^n with n from 1 to infinity) it seems difficult.

Even consider a simpler finite expression. How would you isolate the z^-1 term (that is find the coefficient of z^-1) in this: 4z/(4z-2-z^2) ?

I realize there is a very simple method of doing this problem given by ircdan but I like to see how this method turn out. That is why I posted here instead of the homework section.

 

[benorin]

Suppose that f(z)=\sum_{n=-\infty}^{\infty}a_n(z-a)^n} then the residue of f(z) at z=a is a_-1, the coefficient of 1/(z-a) (it is not the coeff of 1/z for every residue, only for the residue at z=0 is that true).

 

[pivoxa15]

This problem of finding the coefficient of z^-1 in the Laurent series seems impossible by dividing the Laurent series alone. The residue obviously exists but how do you access it? Could it be the case that the z^1 term exists but is impossible to access? The series seems always to start with z and goes upwards in order.

 

[pivoxa15]

But are you using the correct expansion? The Laurent series at 0 tells you nothing about the residues at poles away from zero (which is where all the poles are)

If I expand cos(z) about 0, and use that expansion as my Laurent series, wouldn't that be okay because this series is convergent on the whole plane - including pie/2, which is the first zero. So I can do z/(series of cos(z) about 0)? This is because if I sub pie/2 into z, I would get (pie/2)/0 - which is desired as that shows there is a pole at pie/2.

 

[matt grime]

No, that is not acceptable. Read Benorin's post. The residue at a pole, w, is the coefficient of (z-w)^{-1} in the Laurent expansion *at w* (and stop spelling pi with an e!).

 

[pivoxa15]

I have finally found the series expansion of cos(z) about pi/2.

cos(z) = (1<=n<infinity)sigma((-1)^n(z-pi/2)^(2n-1)/(2n-1)!)

To find Res(f(z),z=pi/2), f(z)=z/cos(z) using a Laurent expansion, I figured long division of z/(series of cos(z) about pi/2). To find the coefficients of 1/(z-pi/2), I would gather all the (z-pi/2)^2 terms (which would be infinitey many) and do z divided by all these terms to get answers/terms which are of the form a/(z-pi/2) where 'a' is an arbitary constant. Is this correct? If so than the trouble with this method is that I will never accumulate all the a/(z-pi/2) terms because there is an inifinite number of them. Although this series (of a/(z-pi/2) terms) should converge because a finite residue exists for this function so I would need to calculate what this series (of a/(z-pi/2) terms) converges to.