# The Complex Logarithm of a Function

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?

• $\gamma$ should start at $z_0,$ I believe. – Thomas Andrews Apr 24 '19 at 17:07
• 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 $k.$ – Thomas Andrews Apr 24 '19 at 17:11
• 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) – Conrad Apr 24 '19 at 17:36

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 $$z$$.