- #1
John O' Meara
- 325
- 0
Experiment with a family of paths with common endpoints, say [tex]z(t) = t + \iota a sin(t) 0 \leq \ t \ \leq \pi[/tex], with real parameter a. Integrate non-analytic functions (Re(z), Re(z^2), etc.) and explore how the result depends on a. Take analytic functions of your choice compare and comment.
Relevant equations:
[tex]\int_C f(z(t))\frac{dz}{dt} dt.\\[/tex]
f(z) =Re(z) = x but x(t)=t; a=1.
[tex]\frac{dz(t)}{dt}=1 +\iota \cos t \ \mbox{therefore} \ \int_C f(z) dz = \int_0 ^{\pi}t(1 + \iota \cos t)dt \\[/tex]
[tex]\mbox{with a=2} \ \int_C f(z)dz = \int_0 ^{\pi} t(\+2\iota \cos t)dt \ \\ \mbox{ with a =3} \int_Cf(z)dz= \ \int_0 ^{\pi} t(1+3\iota \cos t)dt \\[/tex]
We now take a second non-analytic function f(z)=Re(z^2) and a=1 [tex]f(z)=t^2-\sin^2 t \ \int_C f(z)dz \ = \ \int(t^2-\sin^2t)(1+\iota \cos t)dt \\[/tex]
[tex]\mbox{ with a=2} \ \Re{z^2}=t^2-4\sin^2 t \ \mbox{therefore} \ \int_C \Re{z^2}dz \ = \ \int_0 ^{\pi}(t^2-4\sin^2t)(1+2\iota \cos t)dt \\[/tex] .....(v)
Now taking some analytic functions with a=1; [tex]f(z) =z^2 = (t^2+\iota\sin t)^2 \ \ z(t)=t +\iota\sin t \ \ \frac{dz}{dt}=1 + \iota\cos t \ \\ \int_C z^2 dz = \int_0 ^{\pi} (t+\iota\sin t )^2(1+\iota \cos t)dt \\[/tex] ......(vi)
[tex]\mbox{with a=2 } \ z(t)=t+2\iota\sin t \ \ \frac{dz(t)}{dt} \ = \ 1 + 2\iota\cos t \ \ f(z(t)) \ = \ (t \ + \ 2\iota\sin t)^2 \ \ \int_C z^2 dz \ = \ \int_0 ^{\pi} (t+2\iota\sin t)^2(1+2\iota\cos t)dt\\[/tex] ......(vii)
When I carry out the integration for integrals (vi) and (vii) I don't get the same answer for them though they are analytic functions and should be independent of the path. I think the integrals may be wrongly formulated. Thanks.
Relevant equations:
[tex]\int_C f(z(t))\frac{dz}{dt} dt.\\[/tex]
f(z) =Re(z) = x but x(t)=t; a=1.
[tex]\frac{dz(t)}{dt}=1 +\iota \cos t \ \mbox{therefore} \ \int_C f(z) dz = \int_0 ^{\pi}t(1 + \iota \cos t)dt \\[/tex]
[tex]\mbox{with a=2} \ \int_C f(z)dz = \int_0 ^{\pi} t(\+2\iota \cos t)dt \ \\ \mbox{ with a =3} \int_Cf(z)dz= \ \int_0 ^{\pi} t(1+3\iota \cos t)dt \\[/tex]
We now take a second non-analytic function f(z)=Re(z^2) and a=1 [tex]f(z)=t^2-\sin^2 t \ \int_C f(z)dz \ = \ \int(t^2-\sin^2t)(1+\iota \cos t)dt \\[/tex]
[tex]\mbox{ with a=2} \ \Re{z^2}=t^2-4\sin^2 t \ \mbox{therefore} \ \int_C \Re{z^2}dz \ = \ \int_0 ^{\pi}(t^2-4\sin^2t)(1+2\iota \cos t)dt \\[/tex] .....(v)
Now taking some analytic functions with a=1; [tex]f(z) =z^2 = (t^2+\iota\sin t)^2 \ \ z(t)=t +\iota\sin t \ \ \frac{dz}{dt}=1 + \iota\cos t \ \\ \int_C z^2 dz = \int_0 ^{\pi} (t+\iota\sin t )^2(1+\iota \cos t)dt \\[/tex] ......(vi)
[tex]\mbox{with a=2 } \ z(t)=t+2\iota\sin t \ \ \frac{dz(t)}{dt} \ = \ 1 + 2\iota\cos t \ \ f(z(t)) \ = \ (t \ + \ 2\iota\sin t)^2 \ \ \int_C z^2 dz \ = \ \int_0 ^{\pi} (t+2\iota\sin t)^2(1+2\iota\cos t)dt\\[/tex] ......(vii)
When I carry out the integration for integrals (vi) and (vii) I don't get the same answer for them though they are analytic functions and should be independent of the path. I think the integrals may be wrongly formulated. Thanks.