complex closed contour integral

2015 3 12
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更新时间：2022 4 10
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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

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Torsten
Torsten 2015-3-12

1 个投票

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.

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salah zetreni
salah zetreni 2015-3-12
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
Torsten 2015-3-12
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
salah zetreni 2015-3-12
编辑：salah zetreni 2015-3-12
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
Torsten 2015-3-12
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
salah zetreni 2015-3-12
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
Torsten 2015-3-13
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
salah zetreni 2015-3-13
Thank you very much،، I've been mistaken .. Thank you so much for your help
salah zetreni
salah zetreni 2015-3-13
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
salah zetreni 2015-3-13
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
Shariefa Shaik 2017-11-1
explain the procedure sir

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Torsten
Torsten 2015-3-13

0 个投票

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.

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salah zetreni
salah zetreni 2015-3-13
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
Shriya Varanasi 2022-4-10
How do you get the graph?
Torsten
Torsten 2022-4-10
What graph ? The result is the value of the contour integral - thus a single number.

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