2
$\begingroup$

Given an Annulus with $A(0,r,R)$ show by considering Cauchy's Theorem for primitives that there is no holomorphic function with $f'(z)=1/z$

I am struggling to picture this since but it seems like there are issues because $f(z)=log(z)$ isn't well defined in the same range as $1/z$.

Is it worth considering $f(z)$ as anything other than just a function? e.g setting $f(z)=log(z)$ and proving it this way?

  • 0
    "Multi-valued functions" are indeed considered in complex analysis as a way around this very issue, but for the purposes of this exercise $f$ is supposed to be an ordinary, not multi-valued, holomorphic function defined on the annulus. – 2017-02-20

0 Answers 0