Consider the integral ∮dz/(z-a) around |z| = 1 oriented counter-clockwise, where a is any constant such that |a| ≠ 1. Using Cauchy's Theorem and Cauchy's Integral Formula, evaluate this integral for the cases where |a| > 1 and |a| < 1.
I used Cauchy's Integral Formula: f(a) = 1/2πi ∮f(z)dz/(z-a)
since|a|>1 then we don't have to worry about z = a becasue |z| = 1
so i rearranged...
f(a)2πi = ∮f(z)dz/(z-a)
= f(a)2πi = ∮dz/(z-a)
= f(a)2πi = ∮z/(z-a)
= f(a)2πi = ∮[z/(z-a)][1/(z - 0)]dz
this then gives a new function g(z) = z/(z-a) which is analytic in the circle
so now we are doing this for g(0)
= g(0)2πi = ∮[z/(z-a)][1/(z - 0)]dz
for |a| < 1
i just used Cauchy's Integral such that
f(a)2πi = ∮f(z)dz/(z-a) where f(z) = 1 in the given example... so f(a) = 1
sooo... my question is; am I doing this correctly?
Yeppers. One thing only: when you got 0 you assumed, of course, |a| > 1, right?