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I am studying branches of logarithm. I came to know that there are infinitely many branches of logarithm where $\log z = \log |z| + i (\arg z +2k\pi)$, $k \in \mathbb Z$ and $z \neq 0$. Now for each $\alpha \in [0,2\pi)$ if we restrict $\arg z$ to lie inside $(\alpha , \alpha + 2\pi)$ this will yield a branch of logarithm having branch cut $\theta = \alpha$ which is analytic in the cut plane $D_{\alpha} = \mathbb C \setminus \{z \in \mathbb C : z \leq 0 \}$. For each such branch there exists a principal logarithmic function where $k=0$ i.e. $\log z =\log |z| + i \arg_{\alpha} z$ where $z \neq 0$ and $\arg_{\alpha}$ is the restriction of the argument function on $(\alpha,\alpha+2\pi)$ for some $\alpha \in [0,2\pi)$. The principal branch of logarithm corresponds to $k=0$ and $\arg=\arg_{\pi}$ as the argument function which is known as principal argument function.

Now my question is :

"Is the same true for square root?" As we know that $z^{\frac {1} {2}} = \exp (\frac {1} {2} \log z)$. As we know that logarithm has infinitely many branches, each of which is analytic in some certain cut plane. So we can say that $z^{\frac {1} {2}}$ is analytic on a certain cut plane of the corresponding logarithmic branch. But I don't know whether it is analytic on any point on the cut plane of the corresponding logarithmic branch or not!! If it is not so then clearly there are infinitely many branches of square root function. Corresponding to each branch there are two square root functions. One is $z \mapsto |z|^{\frac {1} {2}} e^{\frac {i\arg_{\alpha} z} {2}}$ and the other is $z \mapsto -|z|^{\frac {1} {2}} e^{\frac {i\arg_{\alpha} z} {2}}$ for each $\alpha \in [0,2\pi)$. But for that I need the answer to the question whether $z^{\frac {1} {2}}$ is analytic on the points of the cut plane of the corresponding logarithmic branch or not. If the answer to that question is "no" then only we can extend the concept of logarithmic function to the square root function. I only know that the principal square root function is not continuous on $\mathbb C \setminus \{0 \}$.

Is it true or not? I am in a fix. Please help me.

Thank you in advance.

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The two concepts match. Let us at first revisit the logarithmic function:

The multivalued logarithm is defined as \begin{align*} \log(z)=\log|z|+i\arg(z)+2k\pi i\qquad\qquad k\in\mathbb{Z}\tag{1} \end{align*} In order to make single-valued branches of $\log $ we make a branch cut from $0$ to infinity, the most common being the negative real axis. This way we define the single-valued principal branch or principal value of $\log$ denoted with $\mathrm{Log}$ and argument $\mathrm{Arg}$. We obtain \begin{align*} \mathrm{Log}(z)=\log |z|+i\mathrm{Arg}(z)\qquad\qquad -\pi <\mathrm{Arg}(z)\leq \pi\tag{2} \end{align*}

Now let's look at the square root function:

The two-valued square root is defined as \begin{align*} z^{\frac{1}{2}}&=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)+2k\pi}{2}}\\ &=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)}{2}}(-1)^k\qquad\qquad k\in\mathbb{Z}\tag{3} \end{align*}

In order to make single-valued branches of $z^{\frac{1}{2}}$ we make again a branch cut from $0$ to infinity along the negative real axis. This way we define the single-valued principal branch or principal value of $z^{\frac{1}{2}}$ denoted with $\left[z^{\frac{1}{2}}\right]$ and argument $\mathrm{Arg}$. We obtain

\begin{align*} \left[z^{\frac{1}{2}}\right]&=|z|^{\frac{1}{2}}e^{i\frac{\mathrm{Arg}(z)}{2}}\qquad\qquad -\pi <\mathrm{Arg}(z)\leq \pi\tag{4} \end{align*}

Now we are ready to calculate $e^{\frac{1}{2}\log(z)}$

We obtain from (1) \begin{align*} \color{blue}{e^{\frac{1}{2}\log(z)}}&=e^{\frac{1}{2}\left(\log|z|+i\arg(z)+2k\pi\right)}\\ &=|z|^{\frac{1}{2}}e^{\frac{1}{2}\left(i\arg(z)+2k\pi\right)}\\ &=|z|^{\frac{1}{2}}e^{i\frac{\arg(z)}{2}}(-1)^k\\ &\color{blue}{=z^{\frac{1}{2}}} \end{align*} which coincides with (3).

Taking the principal value $\mathrm{Log}$ we obtain from (2) \begin{align*} \color{blue}{e^{\frac{1}{2}\mathrm{Log}(z)}}&=e^{\frac{1}{2}\left(\log |z|+i\mathrm{Arg}(z)\right)}\\ &=|z|^{\frac{1}{2}}e^{i\mathrm{Arg}(z)}\\ &\color{blue}{=\left[z^{\frac{1}{2}}\right]} \end{align*} which coincides with (4).

We also see the relationship \begin{align*} e^{\frac{1}{2}\log(z)}=\left[z^{\frac{1}{2}}\right](-1)^k \end{align*}

Conclusion: The concepts of logarithm and square root match in the sense that the infinitely many branches of the logarithm yield precisely the two branches of the square root.

Note: This answer is mostly based upon chapter VI from Visual Complex Analysis by T. Needham.

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  • $\begingroup$ Very nice and clean explanation!! $\endgroup$ – Error 404 Jun 15 '18 at 10:40
  • $\begingroup$ @Error404: Thanks. :-) $\endgroup$ – Markus Scheuer Jun 15 '18 at 11:47
  • $\begingroup$ "The concepts of logarithm and square root match in the sense that the infinitely many branches of the logarithm yield precisely the two branches of the square root." I don't see how this makes any sense of the word 'match'. $\endgroup$ – Al Jebr Sep 12 at 19:50
  • $\begingroup$ @AlJebr: Here we have the situation that the exponential function maps the infinite branches of the logarithm to the two branches of the square root function. You might want to consult the reference, which provides helpful information. $\endgroup$ – Markus Scheuer Sep 12 at 20:20

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