solve residue by matlab

plz help me ....

how can I solve this problem by using residue by matlab :

25cos(πz)/((2z+1)^2 (z-2)^2 )

plz help me .. thank u .

Answer 1

[Star Strider]

If you have the Symbolic Math Toolbox, use the partfrac funciton:

syms z
f = 25*cos(pi*z)/((2*z+1)^2 * (z-2)^2 );
pf = partfrac(f, z)

produces:

pf =

cos(pi*z)/(z - 2)^2 - (4*cos(pi*z))/(5*(z - 2)) + (8*cos(pi*z))/(5*(2*z + 1)) + (4*cos(pi*z))/(2*z + 1)^2

[adrian zizo]

thank u ...but I want the value of residue at (z=2 and z=-0.5) how can I find it?

[Star Strider]

The ‘pf’ expression provides you with the partial fraction evaluated at exactly those values. If you substitute them in for ‘z’ in that expression, you will get an infinite result for each, as expected.

[adrian zizo]

but when I enter : syms z

f = 25*cos(pi*z)/((2*z+1)^2 * (z-2)^2 );

pf = partfrac(f, z)

in matlab , matlab give me error ...my matlab version is R2010a

[Star Strider]

According to the online documentation, partfrac was introduced in R2015a, so you wouldn’t have it in R2010a.

I posted the result as ‘pf’ in my Answer, so you have the result of the partfrac operation without having to have the function in your version.

Answer 2

[Konstantinos Sofos]

I would suggest to use L'Hôpital's rule but you could use also the following Adaptive numerical limit (and residue) estimation

[adrian zizo]

I want residue of complex numbers .. plz help me

[Konstantinos Sofos]

Have a look in the link that i attached you before...

for your given example

f = @(z) 25*cos(pi*z)./((2*z+1).^2 .* (z-2).^2 )