complex closed contour integral

plz explain to me how can I use matlab programe for solution of complex contour integral :

example : find ∮((z+1)dz)/(z^3-2z^2 ) around z-2=1 where 1 is Radius of the circle ?

when I solution it Manually the result was equal:-π*i

[Shariefa Shaik]

explain the code sir

Answer 1

[Torsten]

fun=@(z)(z+1)/(z^3-2*z^2);     
g=@(theta)2+cos(theta)+1i*sin(theta);
gprime=@(theta)-sin(theta)+1i*cos(theta);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)

Best wishes

Torsten.

[salah zetreni]

It gives me an error .. see him at the bottom >> fun=@(z)(z+1)/(z^3-2*z^2); g=@(theta)2+cos(theta)+1i*sin(theta); gprime=@(theta)-sin(theta)+1i*cos(theta); q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)

??? Undefined function or method 'integral' for input arguments of type 'function_handle'.

[Torsten]

Try

q1=quad(@(t) fun(g(t)).*gprime(t),0,2*pi)

instead of

q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)

Your MATLAB version does not seem to support "integral".

Best wishes

Torsten.

[salah zetreni]

thank u it worked now ..but it does not give me same results (-π*i),, the result by using matlab it was 0.000+4.7117i

[Torsten]

I get 2*pi*i*3/4 = 3/2*pi*i for the contour integral, and 3/2*pi approximately 4.7117.

Best wishes

Torsten.

[salah zetreni]

yes you get on 3/2*pi*i when the integration around z=1 but in this problem the integration around z-2=1 ..When I dissolve the problem by using cauchy integral I have two poles (at z=0 and at z=2) at z=0 is out the contour it's integration equal zero while at z=2 is inside the contour and it's integration equal -π*i

[Torsten]

Res(f,z=2) = lim(z->2) f(z)*(z-2)=lim(z->2)(z+1)/z^2 = 3/4.

Thus the contour integral of f over |z-2|=1 is equal to

2*pi*i*Res(f,z=2)=3/2*pi*i.

How do you arrive at -pi*i ?

Best wishes

Torsten.

[salah zetreni]

Thank you very much،، I've been mistaken .. Thank you so much for your help

[salah zetreni]

if he asked me on the value of integration around z-1-2J=2 for same problem ,, how can I write function g in matlab ? plz help me .. I can get on gprime

[salah zetreni]

for the same problem he asked me to find a value of integration around z-1-2J=2 by using matlab ... plz see my answer and help me on detect the error :

fun=@(z)(z+1)/(z.^3-2*z.^2);

 g=@(theta)1+2*i+2*cos(theta)+1i*2*sin(theta);

 gprime=@(theta)-2*sin(theta)+1i*2*cos(theta);
 q1=quad(@(t) fun(g(t)).*gprime(t),0,2*pi)
when I do run ,the matlab show me :
Warning: Maximum function count exceeded; singularity likely. 
 In quad at 106

q1 =

-0.0256 + 0.0013i

[Shariefa Shaik]

explain the procedure sir

Answer 2

[Torsten]

What about

fun=@(z)(z+1)./(z.^3-2*z.^2);
g=@(theta)(1+2*cos(theta))+1i*(2+2*sin(theta));
gprime=@(theta)-2*sin(theta)+1i*2*cos(theta);
q1=quad(@(t) fun(g(t)).*gprime(t),0,2*pi)

Best wishes

Torsten.

[salah zetreni]

The result tends to zero ...

q1 =

1.8779e-007 -1.4746e-007i
I think it's correct,, because the actual answer is zero when I use cauchy integral or residue ... thank u veru much

[Shriya Varanasi]

How do you get the graph?

[Torsten]

What graph ? The result is the value of the contour integral - thus a single number.