How to understand $\log{f(z)}$?

Question

For example, let $\Omega$ be the region $Re(z)>1$ which is simply connected, and let $f(z)=z^9$. I want to find an explicit formula for $\log{f(z)}$ such that $\log{f(z)}$ is holomorphic on $\Omega$ and that $\log{f(z)}$ coincides with $\log{x^9}$ for all real $x>1$.

Here is how I did it:

• Since $f$ is holomorphic and nonvanishing on $\Omega$, there is a holomorphic function $g(z)$ on $\Omega$ such that $e^{g(z)}=f(z)=z^9$ and that $g(x)=9\log{x}$ for real $x>1$.

• Similarly, there is a holomorphic function $h(z)$ on $\Omega$ such that $e^{h(z)}=z$ and that $h(x)=\log{x}$ for real $x>1$.

• Since for all real $x>1$ we have $g(x)-9h(x)=0$, we can say that $g(z)-9h(z)=0$ on $\Omega$.

• On the other hand, we can write $h(z)=\log{|z|}+i\arg{z}$ on $\Omega$ where $\arg{z}\in(-\pi,\pi)$.

• Thus, since $g(z)=9h(z)$ on $\Omega$, we have $g(z)=9\log{|z|}+9i\arg{z}$ where $\arg{z}\in (-\pi,\pi)$.

Is my argument correct or not? If it is, then how can I see the formula for $g(z)$ directly? Moreover, if $f(z)$ is not as simple as a polynomial, how can I give the formula for $\log{f(z)}$ such that $\log{f(z)}$ is holomorphic?

Thank you very much.

Answer 1

Your argument is correct. A more concise way of phrasing it (not really that different from what you wrote) is to say that the real part of $\log f(z)$ must be always be the real value $\log |f(z)|$, so choosing a holomorphic branch of $\log f(z)$ amounts to choosing a continuous imaginary part, which is $\arg z$. The constraint that we coincide with the real $\log x^9$ for $x>1$ means that we must choose 0 for $\arg x$, where $x$ is positive real. Then all other choices of $\arg z$ for Re $z>1$ are forced on us.