Bacterius said:Note that in the first case the function (which is $\frac{1}{z}$) is holomorphic everywhere inside the circle of radius $1$ centered at $2i$, since it has a single pole at $z = 0$ which is outside this circle. What does that imply?
In the second case, the pole is inside the circle. What does Cauchy's integral theorem say in this case?
aruwin said:What does having a pole at z=0 mean? By pole you mean the y-axis?
Bacterius said:The function is not holomorphic at z = 0 : there is a "discontinuity" there, but since we're in the complex plane we call these types of discontinuities poles (since it's not really a discontinuity, more of a hole in the domain of the function). See Pole (complex analysis) - Wikipedia, the free encyclopedia
Now if a function is holomorphic everywhere inside a closed convex curve, what does Cauchy's theorem tell you about the integral? (for question 1)
aruwin said:That the integral of f(z)=0? Is that right?
Bacterius said:That's correct. Cauchy's theorem for convex regions states that if $f$ is a function holomorphic in a convex region $C$ and $\gamma$ is a closed curve in $C$, then:
$$\int_\gamma f(z) ~ \mathrm{d} z = 0$$
Now for the second question note that the function isn't holomorphic inside the circle, because the circle contains zero. So you have to use the more general theorem (the integral theoem). First cite the theorem and see how it applies to your integral.
aruwin said:Is this correct for no.2? And by the way, the answer for the first one is just zero, right? But I don't understand that theorem. Why does the integral become zero when f(z) is analytic everywhere within a simply-connected region?
Prove It said:Proof of Cauchy's Theorem
As for the second, I think what you have done is correct, however I would use the Residue Theorem.