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For an analytic function $f$ that does not vanish on a simply connected region, we may define its logarithm to be the function:

$$\log f=g(z):=\int_{y}\frac{f'}{f}dz+c_0.$$

Where $\gamma$ is some path starting at an arbitrary point in the region, and ending at $z$; while $c_0$ satisfies $e^{c_0}=f(z_0)$.

I believe that this logarithm should satisfy under certain conditions that: $$\log f=\log |f|+iarg(f).$$

Am I right, or this is too difficult in general?

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  • $\begingroup$ $\gamma$ should start at $z_0,$ I believe. $\endgroup$ – Thomas Andrews Apr 24 '19 at 17:07
  • $\begingroup$ Depending on how you choose $c_0$, I believe you can only get $\log f = \log |f| + i(\operatorname{arg}(f)+2\pi k)$ for integers $k.$ $\endgroup$ – Thomas Andrews Apr 24 '19 at 17:11
  • $\begingroup$ there is a bit of confusion here when you talk about $\arg$ in the sense that for each $z$, the equation $\log f(z)=\log |f(z)|+ i\arg(f(z))$ picks a value from the infinite set $Arg (c)$, where $c=f(z)$ in a consistent way that makes the function $\arg f(z)=\Im{\log f(z)}$ harmonic (also continuos, real analytic etc) $\endgroup$ – Conrad Apr 24 '19 at 17:36
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The function $g$ satisfies $g' = \frac{f'}{f}$ in the given domain, so that $$ (f e^{-g} )' = f' e^{-g} - f g' e^{-g} = 0 \\ \implies f e^{-g} = \text{const} = f(z_0) e^{-g(z_0)} = f(z_0) e^{-c_0} = 1 \, . $$ Therefore $e^g = f$, i.e. $g$ is “a holomorphic logarithm” of $f$ in the domain. In particular $$ f(z) = e^{g(z)} = e^{\operatorname{Re} g(z)} e^{ i \operatorname{Im}g(z)} $$ which implies that $$ |f(z)| = e^{\operatorname{Re}g(z)} \implies \operatorname{Re}g(z) = \log |f(z)| $$ and that $ \operatorname{Im}g(z)$ is an argument of $f(z)$. So $$ g(z) = \log |f(z)| + i \operatorname{arg}f(z) $$ in the sense that $\operatorname{arg}f(z)$ is a continuous function which is an argument of $f(z)$ for each $z$.

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