# I Finding a branch of a function in the unit disk

1. Dec 15, 2016

If trying to find a branch of $(z^2-1)^{1/2}$, it can be shown that one acceptable answer is: $i e^{0.5 Log(1-z^2)}$

But I just want to clarify, is not the following an acceptable answer, too: $iz e^{0.5 Log (\frac{1}{z^2} -1)}$

It appears the argument of Log in both cases is always defined when inside the unit circle. Are they not both valid branches?

(Where Log denotes the principal branch with a cut on the non-positive real axis.)

Last edited: Dec 15, 2016
2. Dec 16, 2016

### chiro

3. Dec 16, 2016

### RockyMarciano

What definition of branch are you using? Just make sure that the function meets the requirements of the definition. One of the requirements is that for each point z in the valid domain, its value in the branch be a value in the original multivalued function , do you think the second expression meets this condition?

4. Dec 16, 2016

I believe so, no? The only point that would cause trouble is $z = 0$, but when you expand the terms out into its appropriate Laurent series, you have:

$$\lim_{z\to 0} iz e^{0.5 Log (\frac{1}{z^2} -1)} = \lim_{z\to 0} iz \sum_{j=0}^{\infty} [0.5 Log (\frac{1}{z^2} -1)]^j / j!$$

converges since each individual term can be expressed as

$$\lim_{z\to 0} \frac{iz [0.5 Log (\frac{1}{z^2} -1)]^j}{j!} \\ = \lim_{z\to 0} i\frac {[0.5 Log (\frac{1}{z^2} -1)]^j}{(1/z)j!} \\ \stackrel{\text{H}}{=} \lim_{z\to 0} i\frac {[0.5 Log (\frac{1}{z^2} -1)]^{j-1}}{(j-1)!(-1/z^2)(0.5)(\frac{1}{z^2} -1)(-2/z^3))}$$

Where after applying L'Hospital's rule $j-1$ more times, the series would be found to converge. Is there something wrong with this interpretation and why it would be an invalid branch?

Last edited: Dec 16, 2016
5. Dec 16, 2016

### RockyMarciano

You show that function is analytic there, but 0 was the branch point in $(z^2-1)^{1/2}$, so it can't be a branch of the original multifunction if admits it as value.

6. Dec 16, 2016

Perhaps I am misunderstanding a concept, but I have only shown that the infinite series converges as $z\to 0$; $0 + 0i$ is no longer the branch point since I am trying to define the function inside the unit circle for both branches, $F = i e^{0.5 Log(1-z^2)}$ and $G = iz e^{0.5 Log (\frac{1}{z^2} -1)}$. All that was necessary was to construct a branch of $(z^2 - 1)^{1/2}$ such that it is single-valued when inside the unit circle, $|z| < 1$, no?

Thus considering the arguments of Log in each case, for $F$:

$$1 - z^2 > 0 \implies 1 > z^2 \implies 1 > |z|$$

as needed. And for $G$:

$$\frac{1}{z^2} -1 > 0 \implies \frac{1}{z^2} > 1 \implies 1 > z^2 \implies 1 > |z|$$ as well.

Last edited: Dec 16, 2016
7. Dec 16, 2016

### RockyMarciano

Sure, maybe my wording confused you, G is analytic in its domain, what I'm saying is that you should be sure that F and G are both branches of the same multivariable function and not two different functions. Just check they have the same domains.

8. Dec 16, 2016

Which they are, no? They are both easily derived from $(z^2-1)^{1/2}$, and if I plug in arbitrary points (e.g. z = 0.5), then I get:

$$F(0.5) = ie^{0.5Log(0.75)} = i\sqrt{3}/2$$

and

$$G(0.5) = i(0.5)e^{0.5Log(3)} = i(0.5)(\sqrt{3}) = i\sqrt{3}/2$$

Perhaps I am still missing something, but is it not evident $F$ and $G$ are both valid branches to $(z^2-1)^{1/2}$? They are analytic and single-valued in the unit circle, which is the domain of interest for this problem.

9. Dec 17, 2016

### aheight

I believe you would benefit if you learned how to plot them and then visually take them apart graphically to see exactly what is going on when you excise a particular part of a multi-valued function as a "branch" and then compare the plot with the analytic expression you have created for that branch and then test your hypothesis: pick a point on your branch, compute it analytically using your derived expression for the branch and then plot the point. Does the point end up on your branch or does it lay on another section of the function? Continue analyzing the function this way as you cultivate an intuitive understanding of branches.

Last edited: Dec 17, 2016
10. Dec 19, 2016

### RockyMarciano

Yes, it is evident once one does the work of checking the branch cuts for the branches one wants to consider, and then one wouldn't have to ask if both branches are valid in the first place.
Since that was your intial question I assumed you had found that there was some problem with the branches and thus my questions.

In fact the branch cut of F is a subset of the branch cut of G , and certainly they are inside the unit disk.
Now it would be interesting to know why you had doubts about G being a valid branch.

Last edited: Dec 19, 2016
11. Dec 19, 2016

Because I was told G was not a valid branch by my professor and for this lost about 5% of my mark. I don't mind losing the marks, but received an unsatisfactory explanation for why it's an invalid branch in the unit circle so I was hoping to look for alternative explanations.

12. Dec 19, 2016

### RockyMarciano

Ok, hold on a second. Have you checked that the branch cut for $e^{0.5 Log (\frac{1}{z^2} -1)}$ is $(-∞,-1]∪[1,∞)$?, I think it might leave out the ±1point, and therefore not be a valid branch cut, it would not be a problem with the argument of Log but with the branch cut not including points that give you trouble with the root. Just by inspection the problem might also have to do with the imaginary axis being a branch cut, check it out.
Branch finding with non completely trivial functions is such a tedious task anyway, and one is always unsure if some point got mixed in the different brackets of the cut intervals so just follow aheight advice and you'll be fine.

Last edited: Dec 19, 2016
13. Dec 20, 2016

You are right, the issue is what I said about the imaginary axis, the principal branch of $\log z^2$ has a branch cut from 0 to -infinity so any pure imaginary $z$ is not valid.