# understanding a theorem in complex analysis

Here is a text from the book Complex Variables and Applications by Churchill :

The proof to this theorem in the book is understandable except for that the proof and the theorem both doesn't say that each of the integrals on $C$ and $C_k$'s are separately zero also since $f(z)$ is analytic. Why is that so?

• None of those integrals are zero, because $f$ is not analytic in the interior of any of those curves. Look again: $f$ is analytic in the region interior to $C$ and exterior to the $C_j$. – David C. Ullrich Mar 26 '16 at 15:01
• @DavidC.Ullrich - but for example $f(z)=1/{z^2}$ is not analytic in the origin so why its integral around the origin is zero? – Liebe Mar 26 '16 at 15:04
• When I said none of those integrals are zero of course I meant they're not necessarily zero. Think about $1/z$ instead of $1/z^2$. – David C. Ullrich Mar 26 '16 at 15:08
• @DavidC.Ullrich - yes I got what you mean. But I really don't understand why integral of $f(z)=1/{z^2}$ is zero while $f(z)$ is not analytic? PS It's a bit different question but still relevant matter and I asked it prev and it marked duplicate but it wasn't at all. – Liebe Mar 26 '16 at 15:12
• Cauchy's Theorem says if blah blah blah then the integral is zero. That's "if", not "if and only if". The integral of $1/z^2$ about the unit circle is zero because it is. (This may be clearer later, when you know the Residue Theorem.) – David C. Ullrich Mar 26 '16 at 15:25