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Suppose I have a branch of the logarithm, that is, a continuous function $L(z)$ on some region $\Omega$ such that $e^{L(z)} = z$. We see that this defines a branch for the square root function on $\Omega$, via $\sqrt{z} = \exp(1/2 L(z))$, since

$$(\exp(1/2 L(z))^2 = \exp(L(z)) = z$$

I am wondering if a sort of converse of this holds. Suppose on the other hand, we have a branch for the square root, i.e. some continuous function $R(z)$ on $\Omega$ such that $R(z)^2 = z$. Is there some way to get a branch of the logarithm from $R(z)$? If so, does this generalize (i.e. what branches for multi-valued functions will determine a branch of the logarithm)?

user091823
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3 Answers3

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While every cut for the square-root function is admissable as a cut for the logarithm in $\mathbb{C}\backslash\{0\}$, the complex plane without the origin, there are only two "layers" of the square-root function and countably many for the logarithm. So it cannot be said that a branch of the square-root function "determines" a branch of the logarithm, unless you are willing to simply impose some choice of branches ad hoc.

hardmath
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Suppose $\Omega=\{z\in\mathbb{C}:z\nleq 0\}$. For $z\in\Omega$, let $\mathrm{Arg}(z)$ indicate the unique $\theta\in(-\pi,\pi)$ such that $z=|z|e^{i\theta}$. Define $R(z)=|z|^{1/2}e^{\frac{1}{2}\mathrm{Arg}(z)}$. This is a branch of the square root.

For each $k\in\mathbb{Z}$, $z\in\Omega$, let $L_k(z)=\log|z|+i(\mathrm{Arg}(z) + 4\pi k)$. Each $L_k$ is a distinct branch of the logarithm, but each one should satisfy $R(z)=\mathrm{exp}(\frac{1}{2}L_k(z))$.

Cameron Buie
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  • As Christian (pertinently) points out below, "some way to get a branch of the logarithm" is vague. You gave a fairly natural convention typically used to obtain an $n$th root branch from a branch of the logarithm, but without some notion of the constaints you desire of the "converse" it is difficult to answer your question definitively.

    Above, I attempt to apply an analogous constraint to the converse, to suggest why it isn't sufficient (in and of itself), and that more restrictions are required if you are to have any hope of a definitive answer.

     – Cameron Buie Apr 21 '12 at 09:56
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A region $\Omega\subset{\mathbb C}$ admits a continuous branch $R$ of $z\mapsto\sqrt{z}\ $ iff $\ \Omega$ contains no closed path around the origin, and the same is true for the existence of a continuous branch ${\rm Log}$ of $z\mapsto\log z$.

Now given such an $\Omega$ you needed the exponential function to create $R$ from ${\rm Log}$. If you want us to "create" ${\rm Log}$ from $R$ you would have to indicate somehow what the allowable means are. Note that even in the real domain there is no natural process of this sort.

Christian Blatter
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