Is $\log(z-\alpha)$ well defined

complex-analysislogarithms

Could you help me with the following problem please:

Show that if $U\subset{\mathbb{C}}$ is simply connected domain and $\alpha \in \mathbb{C}\setminus U$, then log$(z-\alpha)$ is well defined for all $z\in U$

I have no idea how to solve this problem, it arose at the time of proving the Riemann mapping theorem, but it does not occur to me how to do the proof, also, I have tried to see it taking the case in which U is an open ball, but not I can understand well what the exercise says, since if, for example, I took the unit ball centered on the origin, there would still be values in which the logarithm would not be well defined. I hope you can help me with this, thank you.

My try:

We fix a point $z_0\in U$ and let's define:
$$
\log(z-\alpha)=\displaystyle\int_{\gamma(z)}\dfrac{1}{w-\alpha}dw+\log(z_0-\alpha)
$$
with $\gamma(z)$ a path in $U$ that joins $z$ with $z_0$.

And it would be necessary to see that in this case the integral does not depend on the path, I think that would suffice, but I have not managed to prove it, it occurs to me that it is because it is a simply connected domain and applying Cauchy's theorem

Best Answer

I think you are on the right track. I'd proceed by doing a further step remembering the following facts:

  1. Cauchy's integral theorem: if $f: U\to \Bbb C$ is a holomorphic function on the domain $U$, then $$ \oint\limits_\gamma f(w)\mathrm{d}w=0 $$ for each simple closed curve contained in $U$.
  2. If $\alpha \in \Bbb C\setminus U$ then $$ f(w)=\frac{1}{w-\alpha}\quad \forall w\in U $$ is holomorphic on the whole $U$. To see this note that, by putting $w\equiv x+iy$, we have $$ \begin{split} \frac{\partial}{\partial x} f(x,y) &= \frac{\partial}{\partial x} \frac{1}{x+iy-\alpha} \\ & = -\frac{1}{(x+iy-\alpha)^2}\\ & = -i\frac{\partial}{\partial y} \frac{1}{x+iy-\alpha}= -i\frac{\partial}{\partial y} f(x,y) \end{split} $$ thus $f$ satisfies the Cauchy-Riemann equations on the whole $U$ since it is continuous with its partial derivatives for any $w\in U$ due to the hypothesis $\alpha \in \Bbb C\setminus U$.

Now the definition of $\log(z-\alpha)$ as given above is seen to be path independent since, for any two continuous paths $\gamma_1[z],\gamma_2[z]: [0,1]\to U$ joining $z_0$ to $z$ in $U$ $$ \oint\limits_{\gamma_1[z]}\frac{1}{w-\alpha}\mathrm{d}w = \oint\limits_{\gamma_2[z]}\frac{1}{w-\alpha}\mathrm{d}w \iff \oint\limits_{\gamma_1\cup [-\gamma_2]}\frac{1}{w-\alpha}\mathrm{d}w=0 $$ provided having defined the closed curve $\gamma_1\cup [-\gamma_2]:[0,1]\to \Bbb C$ as $$ \gamma_1\cup [-\gamma_2](t)= \begin{cases} \gamma_1[z](2t) & t\in \big[0,\frac{1}{2}\big[\\ \gamma_2[z](2- 2t) & t\in \big[\frac{1}{2},1\big] \end{cases} $$ Thus $\log(z-\alpha)$ is univocally defined.

Final notes

  • As @Haus said in the comments, the choice of $z_0\in U$ is exactly the choice of the branch of the logarithm we are willing to use: once you chose $z_0$, the additive constant becomes $\log(z_0-\alpha)$ and is fully determined for the whole simply connected domain $U$. If we consider $\alpha=0$ and $U=\Bbb C\setminus {0}$ (which is not simply connected) we see that it is not so: depending on how we encircle $0$ by the path $\gamma[z]$ (clockwise or counterclockwise, starting from an arbitrary $z_0\neq 0$), we have that $$ \log z =\log |z| + i\arg z $$ where the imaginary part is only known modulo $2n\pi i$ for all $n\in\Bbb N$ and thus not uniquely determined.
  • The curve $\gamma_1\cup[-\gamma_2]$ has been defined in perhaps a cumbersome way in order to show how it is closed. Intuitively speaking, when the parameter $t$ rises from $0$ to $\frac{1}{2}$, the mapped point "moves" along $\gamma_1$ from $z_0$ to the (chosen and fixed) $z$: when it further rises beyond $\frac{1}{2}$ to $1$, the mapped point "moves" from $z$ to $z_0$ "reversing its motion". This later behavior is the reason for the highly nonstandard and perhaps unfortunate notation $[-\gamma_2]$ I've chosen for this part of the composite path: I wanted to find something that would remind the reader of the addition of simplexes in euclidean spaces, in order to make the concept intuitively clear, but I've obviously failed.
  • Strictly speaking, the reasoning above requires the two paths $\gamma_1[z]$ and $\gamma_2[z]$ to be non intersecting: however, the integral calculated on them can be seen to be equal to the one calculated on an auxiliary path joining $z$ to $z_0$ and not interesting any of them, thus the independence of the definition of the complex logarithm holds without any restriction on the used path.
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