# complex main branch of a logarithmic function holomorphic correct

Let $z:= x+iy$ It will now be shown that

$$f(z)= f(x+iy) = \frac{1}{2} \log(x^2 + y^2) + i \arctan\left(\frac{y}{x}\right); (z\in \mathbb{C}, x = \operatorname{Re} z \ne 0),$$

where $arctan$ denotes the main branch of the real $\arctan$ function and $\log$ the real logarithmic function, is (1) : a holomorphic function $(f(z)\in \mathcal{O}(G))$ and that it is (2): a logarithmic function, and also the main branch of the logarihmic function.

For (1): The Cauchy Riemann conditions are checked. $u_{x}=\frac{x}{x^{2}+y^{2}} = v_{y}$ and $u_{y} = \frac{-y^{2}}{x^{2}+y^{2}} = -v_{x}$. Thus f(z) is holomorphic.

(2): A logarithmic function must satisfy $\exp(\log(z)) = z = re^{i\phi}$ And we see that with $\frac{1}{2}\log(x^2+y^2) = \log((x^{2}+y^{2})^{1/2})$ :

$$\exp(\log((x^2 + y^2)^{1/2}) + i\arctan(y/x)) = \sqrt{x^2+y^2}e^{i\arctan(y/x)} = |z|e^{i\phi}$$

and thus $f(z)$ must be a logarithmic function. Since only the main branch of the arctan function is allowed, that means that this is also the main branch of the complex logarithm.

It wasn't shown here that $\arctan(y/x) = \phi$, how do I do that? Is it true that $f(z)$ is a logarithmic function only in the right half plane ?

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