2011-06-02, 11:07 PM
∮1/z dz = ∫ [-sin(θ
+ icos(θ]/[cos(θ + isin(θ] dθ, [0,2π]
= ln(cos(θ + isin(θ), [0,2π]
= ln(e^(iθ), [0,2π]
= iθ, [0,2π]
= 2πi
Okay. So, how would I correct for this in the other two paths? Let's just say that in general:
∮1/z dz = ln(z), [0,2π],
where z has a parametrization in the form of z(e^(iθ). Since, the parametrization for the second path is:
z = 2cos(θ+isin(θ
= 3/2 e^(iθ + 1/2 e^(-iθ
What am I supposed to do to cancel out the natural log in the following?
ln(3/2 e^(iθ + 1/2 e^(-iθ), [0,2π]
Using sines and cosines isn't any more revealing, and path 3 is un-parametrizable with in the form of e^(iθ, so how do I proceed?
+ icos(θ]/[cos(θ + isin(θ] dθ, [0,2π]= ln(cos(θ
+ isin(θ), [0,2π]= ln(e^(iθ
), [0,2π]= iθ, [0,2π]
= 2πi
Okay. So, how would I correct for this in the other two paths? Let's just say that in general:
∮1/z dz = ln(z), [0,2π],
where z has a parametrization in the form of z(e^(iθ
). Since, the parametrization for the second path is:z = 2cos(θ
+isin(θ= 3/2 e^(iθ
+ 1/2 e^(-iθWhat am I supposed to do to cancel out the natural log in the following?
ln(3/2 e^(iθ
+ 1/2 e^(-iθ), [0,2π]Using sines and cosines isn't any more revealing, and path 3 is un-parametrizable with in the form of e^(iθ
, so how do I proceed?