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

= 0/(0-a)

=0


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

so f(a)2πi

= 2πi

sooo... my question is; am I doing this correctly?

Yeppers. One thing only: when you got 0 you assumed, of course, |a| > 1, right?

yep i was doing |a| > 1

thanks for the input. i thought i was on the right track