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

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 $\dfrac{1}{z}$.

Am I looking to show that $\int_Ag(z)dz = 0$ where $g(z) = \dfrac{1}{z}$?

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Let $C$ be the circle of radius $1$ and center $(0,0)$, one has: $$\int_C\frac{\mathrm{d}z}{z}=\int_0^1\frac{2i\pi e^{2i\pi t}}{e^{2i\pi t}}\,\mathrm{d}t=2i\pi\neq 0.$$ If $z\mapsto 1/z$ has a primitive in $A(0,1,2)$, the above integral must be $0$. Whence the result.

Remark. I let you adapt the proof for $A(0,r,R)$ when $r<R$.

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  • $\begingroup$ If $g(z) = 1/z$ had a primitive $G(z)$ then $\int_C g(z) dz = \int_1^{e^{2i\pi}} g(z)dz= G(e^{2i \pi})-G(1)=0$ $\endgroup$
    – reuns
    Commented Feb 19, 2017 at 20:25

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