Question

我试图利用综合指挥在几个循环封闭道路上找到复杂的线/构件。我关于第1/(z-i)^2圈子({z : > ><2}穿透器)的代码一度为:

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

(我预计答案为0,matlab提供6.6613*10^(-16)-4.4409*10^(-16)i)。

My code for the integral of e^z/(z(z^2-9)) over the circle {z:|z-2|=3} traversed once anticlockwise is as follows:

fun = @(z) exp(z) ./(z .* (z.^2-9));
g = @(t) 2+3 .*(cos(t) + 1i .* sin(t));
gprime = @(t) 2+3 .*(-sin(t) + 1i .* cos(t));
q1 = integral(@(t) fun(g(t)) .* gprime(t),0,2 .* pi)

(我预计答案为pi/9(e^3-2)i,但书记处提供5.431+6.3130i)。

As can be seen above, my problem is that while the code gives accurate values when the circular path is centred at the origin, it fails otherwise; sometimes giving an accurate imaginary part but inaccurate real part or just a completely inaccurate answer.

Can anyone see what is going wrong?

Answer 1

我在评论中回答了第一个问题。

关于第二个问题,你在计算衍生工具时犯了一个错误:重复不变2应当消失。这样,你将获得6.3130,完全符合理论价值。

fun = @(z) exp(z) ./(z .* (z.^2-9));
g = @(t) 2+3 .*(cos(t) + 1i .* sin(t));
gprime = @(t) 3 .*(-sin(t) + 1i .* cos(t));
q1 = integral(@(t) fun(g(t)) .* gprime(t),0,2 .* pi)

I take the opportunity to advise you, instead of the previous method, to compute complex integrals by "way points" method (see for example https://uk.mathworks.com/help/matlab/math/complex-line-integrals.html) ; here, it would be

C=[-2+i,-2-i,5-i,5+i];
integral(@(z) (exp(z) ./(z .* (z.^2-9))),1,1, WayPoints ,C)

如果C是一平方(更概括地说是一平方),则只包含所希望的极地。 (途径2和3参数为1,1)

附录:按残余物对理论价值的快速个人检查:

2i pi(Res(f,0)+Res(f,3)=2i pi(1/(-9)+e^3/(27-9))。

Answer 2

您对第二个问题的回答是不正确的。数学告诉我,答案应当是:

pi*(2/45)*(5 + (exp(2))*(5*exp(1) - 9)) + (pi/9 *(exp(3) - 2))*i

= 5.435120011473026 + 6.313043326012592i

这几乎是你从马特拉布获得的答案。

请注意,Matlab的integral功能履行全数>集成,因此,你获得的结果将出现机对冲错。这意味着,你对第一个问题的结果基本上也是正确的;这或许是你在数字融合方面的最佳答案。

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