What are the branches of the square root function?

Solution 1:

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.