∫ (z')^n dz over a curve, C : |z| = R
Where z' is the conjugate of z (if z = x + iy then z' = x - iy)
and C is a circle of radius R centered at the origin
I need to evaluate the integral when
a) n ≠ 1
b) n = 1
I know the answers are a) 0 and b) 2πiR² but I don't know how to do this, since a function with z' in it is not holomorphic.
Where z' is the conjugate of z (if z = x + iy then z' = x - iy)
and C is a circle of radius R centered at the origin
I need to evaluate the integral when
a) n ≠ 1
b) n = 1
I know the answers are a) 0 and b) 2πiR² but I don't know how to do this, since a function with z' in it is not holomorphic.
-
Parameterize C by z = Re^(it), dz = iRe^(it) dt.
Remember that (e^(it))' = e^(-it), with ' denoting conjugation.
Why: (e^(it))' = (cos t + i sin t)' = cos t - i sin t = e^(-it).
------------------
So, ∫c (z')^n dz
= ∫(t = 0 to 2π) (Re^(-it))^n * iRe^(it) dt
= iR² * ∫(t = 0 to 2π) e^((1-n)it) dt.
If n = 1, then we have
iR² * ∫(t = 0 to 2π) 1 dt = 2πiR².
Otherwise, we have
iR² * e^((1-n)it) / (i(1 - n)) {for t = 0 to 2π} = 0.
I hope this helps!
Remember that (e^(it))' = e^(-it), with ' denoting conjugation.
Why: (e^(it))' = (cos t + i sin t)' = cos t - i sin t = e^(-it).
------------------
So, ∫c (z')^n dz
= ∫(t = 0 to 2π) (Re^(-it))^n * iRe^(it) dt
= iR² * ∫(t = 0 to 2π) e^((1-n)it) dt.
If n = 1, then we have
iR² * ∫(t = 0 to 2π) 1 dt = 2πiR².
Otherwise, we have
iR² * e^((1-n)it) / (i(1 - n)) {for t = 0 to 2π} = 0.
I hope this helps!