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I am currently studying about Branch points and Branch cuts. I think, I understood the definition of these two concepts and I can find Branch points and Branch cuts of some functions.

For example:
$z\to\sqrt{z(1-z)}$ has two branch points $0$ and $1.$ Because as we travel along a small circle around $0$ or $1,$ one time, argument of the function changes to $0\to \pm\pi i.$
On the other hand, $z\to\sqrt{z}+\sqrt{1-z}$ has three branch points $0, 1$ and $\infty.$
(Correct me if I am wrong.)

Here my question is:

There are identities which fails on some branches of logarithm.
Is there any way to determine the region for which these familiar identities valid using Branch points and Branch cuts?

For example: $$\color{Green}{\sqrt{z-1}\sqrt{z+1}=\sqrt{z^2-1}}$$ does hot holds for $z=-2,$ if we choose $\sqrt{-1}=i.$

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  • $\begingroup$ If $ab \ne 0$ then $\log(ab) = \log(a)+\log(b)+2i k \pi$ for some $k \in \mathbb{Z}$ is always true for any branch of $\log(z)$. Changing of branch means changing $k$ (which depends in general on $a,b$) $\endgroup$
    – reuns
    Jan 8, 2017 at 20:28
  • $\begingroup$ Thank you @user1952009 Yes. I got your idea. But how can it helps to solve this problem? $\endgroup$
    – Bumblebee
    Jan 10, 2017 at 4:35

1 Answer 1

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In THIS ANSWER, I discussed the meaning of the identity

$$\log(z_1z_2)=\log(z_1)+\log(z_2) \tag 1$$

In that expression, the equality is interpreted as a set equality. This means for any value of $\log(z_1z_2)$ can be expressed as the sum of some value of $\log(z_1)$ and some value of $\log(z_2)$. In addition, the sum of any values of $\log(z_1)$ and $\log(z_2)$ can be expressed as some value of $\log(z_1z_2)$.


Now, suppose $f(z)=\sqrt{z^2-1}=\sqrt{(z-1)(z+1)}$. By definition, we have

$$\begin{align} f(z)&=e^{\frac12\log((z-1)(z+1))}\\\\ &=e^{\frac12\left(\log(z-1)+\log(z+1)\right)}\\\\ &=e^{\frac12\sqrt{z-1}}e^{\frac12\sqrt{z+1}}\\\\ \sqrt{z^2-1}&=\sqrt{z-1}\sqrt{z+1}\tag2 \end{align}$$

The equality in $(2)$ is a set equivalence analogous with $(1)$.


EXAMPLE:

For the example given in the OP, $z=-2$. We denote by $z_1$ and $z_2$, $z_1=z+1$ and $z_2=z-1$. Clearly, $z_1=-1$, $z_2=-3$, and $z_1z_2=3$.

The multi-valued term $\log(z_1z_2)$ is given by

$$\log(z_1z_2)=\log(3)=\log(|3|)+i2n\pi\tag 3$$

for any integer $n$. If we define $\log(z_1)=i\pi$ and $\log(z_2)=\log(|3|)+i\pi$, and if $n=1$ in $(3)$, then $\log(z+1)+\log(z-1)=\log(z^2-1)$. However, for any other $n$, the equality does not hold.

If we use $(3)$ to calculate $\sqrt{z^2-1}=\sqrt{z_1z_2}$, then we obtain

$$\sqrt{z_1z_2}=\sqrt{3}=\sqrt{|3|}e^{in\pi}\tag 4$$

For $n$ odd, the equality $\sqrt{z_1z_3}=-\sqrt{|3|}=\sqrt{z_1}\sqrt{z_2}$ holds, while for $n$ even, the equality does not hold.


Now, let's be more general. We cut the plane using the principal branch for $\sqrt{z-1}$ and observe that

$$\sqrt{z-1}=\sqrt{|z-1|}e^{i\text{Arg}(z-1)/2}$$

where $-\pi<\text{Arg}(z)\le \pi$ is the principal argument of $z$.

Similarly, using the principal branch for $\sqrt{z+1}$, we see that

$$\sqrt{z+1}=\sqrt{|z+1|}e^{i\text{Arg}(z+1)/2}$$

for $-\pi<\text{Arg}(z)\le \pi$.

If we wish to use the principal branch for $\sqrt{z^2-1}$, then the equality $\sqrt{z^2-1}=\sqrt{z-1}\sqrt{z+1}$ does not hold in general. Rather, the equality holds as follows:

If $-\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le \pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)}$.

If $\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le 2\pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)+2\pi}$

If $-2\pi<\text{Arg}(z-1)+\text{Arg}(z+1)\le -\pi$, then $\sqrt{z^2-1}=\sqrt{|z^2-1|}e^{i\text{Arg}(z^2-1)-2\pi}$

More simply expressed, the equality holds by using the expression $\arg(z^2-1)=\text{Arg}(z-1)+\text{Arg}(z+1)$.


Let's use the previous example in which $z=-2$. We find that $\text{Arg}(z-1)=\text{Arg}(z+1)=\pi$ and $\text{Arg}(z^2-1)=0$. Indeed, we have

$$\sqrt{z-1}\sqrt{z+1}=\sqrt{z^2-1}$$

when we take $\arg(z^2-1)=\text{Arg}(z^2-1)+2\pi=0+2\pi=2\pi$.

More simply, we have equality with $\arg(z^2-1)=\text{Arg}(z-1)+\text{Arg}(z+1)=2\pi$.

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  • $\begingroup$ I think, I got your point. Thank you very much. If I found any difficulty related to this, I will ask. $\endgroup$
    – Bumblebee
    Apr 2, 2017 at 21:56
  • $\begingroup$ Pleased to hear. And you're welcome. Feel free to ask anytime. -Mark $\endgroup$
    – Mark Viola
    Apr 2, 2017 at 23:15

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