Complex Analysis of the Log Function

Question

Let the square root be defined by the principal branch of the log function. Compare the function $\sqrt{z^2-1}$ and $\sqrt{z-1} \sqrt{z+1}$. Where are the discontinuities of each function?

My attempt: Let $\sqrt{z^2-1} = e^{\frac{1}{2} (\ln |z^2-1| + i\arg(z^2-1))}$

This function is discontinuous when $z^2-1$ would take values on the negative part of the real line, as the principal branch is $(-\pi,\pi]$.

Consider a $k \in \mathbb{R}$, $k>0$.

If $z^2-1=-k \implies z^2 = -k+1 \implies z = \sqrt{-k+1}$ or $z=-\sqrt{-k+1}$.

Therefore, for $z\in (-1,1)$ and $(-i\infty,i0)\cup(i0,i\infty)$, the given function would be discontinuous.

Similarly, if $z+1=-k \implies z=-1-k$. This implies for $z \in (-\infty,-1)$, $\arg(z-1)$ will be discontinuous. If $z-1=-k \implies z=-k+1$. This implies for $z\in (-\infty,1), \arg(z+1)$ will be discontinuous.

Therefore, for $z\in (-\infty,1)$, $\sqrt{z-1} \sqrt{z+1}$ will be discontinuous.

Is this proof correct? If yes, why does such a decomposition yield a different result?

Answer 1

Always keep in mind that from the mathematical view point, a function is not just a formula but also it requires specifying a domain in which it is meaningful. So in particular for the two formulas that you mention, are not just continuous but meaningful in different domains, and the equality between them can be only asserted in the intersection of them! (i.e.: outside the negative real axis)

The second expression is simply not well-defined in the negative real axis.

Answer 2

Branch cut for $\boldsymbol{\sqrt{z^2-1}}$

Define $$ g(z)=\int_{\sqrt2}^z\left(\frac1{w-1}+\frac1{w+1}\right)\mathrm{d}w $$ where the path from $\sqrt2$ to $z$ does not cross $[-1,1]$. If the path circles the branch cut at $[-1,1]$, $g(z)=\log\left(z^2-1\right)$ increases by an integer multiple of $4\pi i$; that is, $2\pi i$ times the sum of the residues of the singularities at $z=-1$ and $z=1$. Since $e^{2\pi i}=1$, $$ \sqrt{z^2-1}=\exp\left(\tfrac12g(z)\right) $$ is well-defined with a branch cut at $[-1,1]$.

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Branch cut for $\boldsymbol{\sqrt{z-1}\sqrt{z+1}}$

If we take the branch cut of $\sqrt{z-1}$ to be $(-\infty,1]$ and the branch cut of $\sqrt{z+1}$ to be $(-\infty,-1]$; thus, the branch cut for the product, $\sqrt{z-1}\sqrt{z+1}$, would be $(-\infty,1]$.

However, each of these functions jumps by a multiple of $-1$ across the portion of the branch cuts from $(-\infty,-1)$; thus, their product is continuous across that portion of the branch cut. Just as $\frac{z^2-1}{z-1}$ has a removable singularity at $z=1$, for $\sqrt{z-1}\sqrt{z+1}$, the portion of the branch cut at $(-\infty,-1)$ might also be considered removable, leaving us with the same branch cut as we had for $\sqrt{z^2-1}$; that is, $[-1,1]$.