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From the definition given by wikipedia and Cauchy's theorem i can find the branch points of $\arcsin$ through its derivative $\displaystyle\frac{1}{\sqrt{1-x^2}}$

Are -1 and 1 simple pole of this expression ? (i'm a bit confused because of the fractional power)

Also, there is also a branch point at infinity. How do i find this branch point ? what are the order of all the branch points of arcsin ?

From wikipedia, i know that simple pole of derivative means logarithmic branch point, so there is no order if -1 and 1 are simple pole of $\displaystyle\frac{1}{\sqrt{1-x^2}}$ ?

Geralt of Rivia
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3 Answers3

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Just use the Euler formula $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$. Having $w=\arcsin(z)$ and $\sin(w)=z$ with a bit of algebra gives : $\arcsin(z)=-i\log(iz+(1-z^2)^{\frac{1}{2}})$. Just look at this. Because of the square root you have branch points at $z=\pm 1$, zero is not a branch point here. Infinity is a branch point because: If you substitute $z=\frac{1}{t}$ and look what is going on if $t \to 0$ you would enevetably convince yourself that infinity is a brnch point. i.e you have term of the sort $\log(t)$. So you should have branch cut extending to infinity.

EDIT: There is a long discussion about the branches of $\sqrt{z^2-1}$, following in the comment section, the argument boils down to the $\sqrt{z^2}$. Clearly $$ \sqrt{z^2} = \mp z, $$
this is why this function is multi valued. However, $$ \sqrt{\left(e^{i0}\right)^2}=\sqrt{\left(e^{i2\pi}\right)^2}=1 $$ Or in other words, by going around the origin the function does not change value. Therefore, it is multi-valued but has no branch points! This is the reason why $\sqrt{z^2-1}$ does not have a branch point at infinity, which can be checked by substituting $z=\frac{1}{t}$ and investigating $t\rightarrow 0$ .

Alexander Cska
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  • Hi Alexander. I hope that you are staying safe and healthy. Just curious … How would one evaluate the integral $\oint_{C}\log\left(iz+\sqrt{1-z^2}\right),dz$ where $C$ is any rectifiable curve that encloses (one time) the branch points at $z=-1$ and $z=1$? It seems that $\arg\left(z+\sqrt{z^2-1}\right)$ and $\arg(z)$ lie in the same quadrant when we cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$. Then, on $|z|=r>1$, as $\arg(z)$ goes from $0$ to $2\pi$ the argument of $z+\sqrt{z^2-1}$ does likewise. This would imply that there is a branch point of the logarithm ($z=0$). – Mark Viola May 06 '20 at 15:47
  • Dear @MarkViola, the branch points are determined by $\sqrt{z^2 -1}$ only. $z+i\sqrt{z^2 -1}=0$ has no solution. I hope that I have not misunderstood your the question. I wish you good health and good fortune in these difficult times. Stay safe, stay well! – Alexander Cska May 07 '20 at 13:57
  • Hi Alexander. Thank you for the quick reply. Indeed $z+\sqrt{z^2-1}\ne0$. Yet, when we encircle the branch points at $-1$ and $1$, the argument of $z+\sqrt{z^2-1}$ changes by $2\pi$. Hence, we change branches of the logarithm $\log\left(z+\sqrt{z^2-1}\right)$ – Mark Viola May 07 '20 at 15:06
  • Hi Mark, this is a composition of several functions. Form them only the square root is problematic, the argument of the log is never zero ($\sqrt{1-z^2}+iz \ne 0$). Since the branches extend to Infinity ($\infty$ is a branch point), I don't see how one could encircle the points $ −1,1$ – Alexander Cska May 07 '20 at 22:59
  • Choose the branch cuts from $-1$ to $\infty$ and from $1$ to $\infty$. Note that on $[1,\infty]$, the overlapping cuts render $z+\sqrt{z^2-1}$ continuous across them (only where they overlap). This leaves the "slit" $[-1,1]$. Now, encircling the "slit", the argument of $z+\sqrt{z^2-1}$ goes from $0$ to $2\pi$. Does that make sense now? – Mark Viola May 07 '20 at 23:05
  • Heuristically, think of the branch points at $\pm1$ blending together at $z=0$ as $|z|$ gets "large." Then, $z+\sqrt{z^2-1}\sim 2|z|$. And clearly $\log(2z)$ has a branch point at $0$. So, (again heuristically) $z=0$ is s hidden branch point. The rigorous manifestation of this "hidden" branch point is that the argument of $z+\sqrt{z^2-1}$ goes from $0$ to $2\pi$ as both branch points are encircled (once). – Mark Viola May 07 '20 at 23:21
  • The $z^{1/n}$ exists, only if by going on a closed curve $\gamma$ the argument changes by $kn2\pi$. Where $k,n$ are integers. In this case, using the logarithmic property of the argument $\Delta_{\gamma}(z^2-1)=\Delta_{\gamma}(z-1)+\Delta_{\gamma}(z+1)$. The argument of the root function should also not be equal to zero. You can check this for a given domain just by tracing the argument along the curve. This leads to the branch cuts $(-\infty,-1)\cup(1,\infty)$. I don't see how can these points be encircled. In this particular case, the square root is the only source of multivaluedness. – Alexander Cska May 08 '20 at 09:18
  • Follow tje argument of $z+\sqrt{z^¥-1}$ ariund the slit. It changes by $2\pi$. – Mark Viola May 08 '20 at 14:01
  • The problem is $\sqrt{(z-1)(z+1)}$, using the logarithmic property going around the line segment $-1,1$ along the closed contour $\gamma$ (say $|z|=2$) one gets $\Delta_{\gamma}(z-1)+\Delta_{\gamma}(z-1)=2\pi+2\pi$ which is intact with the principle branch definition for $\sqrt{z}$. – Alexander Cska May 08 '20 at 17:59
  • Note that $(2\pi+2\pi)/2=2\pi$. So, $\sqrt{z^2-1}$ is continuous on $\mathbb{C}\setminus[-1,1]$. But note that its argument increases by $2\pi$, as does the argument of $z$ itself. Then, we are faced with a discontinuity in the logarithm function since we have an increase in $2\pi$ of the argument of $z+\sqrt{z^2-1}$. Check to see that the argument of $z$ and the argument of $\sqrt{z^2-1}$ always lie in the same quadrant when choosing cuts from $1$ to $\infty$ and from $-1$ to $\infty$. Also notice that the residue at infinity does not exist. ...continuing ... – Mark Viola May 08 '20 at 18:29
  • However, note that the function $f(z)=\log\left(z+\sqrt{z^2-1}\right)-\log(z)$ has a residue at infinity of $0$ (zero). And now, as we encircle the slit $[-1,1]$, do not change the argument of $1+\sqrt{1-z^{-2}}$ by $2\pi$. So, the need to cut the plane (other than the slit for the square root) goes away. This means that for any contours $C_1$ and $C_2$ that encircle the slit one time, $\oint_{C_1}f(z),dz=\oint_{C_2}f(z),dz$. And also, it means that $\int_{C_1}\log(z+\sqrt{z^2-1}),dz=\oint_{C_1}\log(z),dz$, and these depend on how we define the branch of $\log(z)$.. – Mark Viola May 08 '20 at 18:31
  • No it is not the change of the argument is $4\pi$ you don't divide by 2. We are dealing with the argument of the function under the square root only. – Alexander Cska May 08 '20 at 18:48
  • After taking the square root the argument gets halved. – Mark Viola May 09 '20 at 03:52
  • No it does not, because it is not computed in this way! Check the definition of the complex $\log{f(z)}$ and $f(z)^{\frac{1}{n}}$ on a simply connected domain. We are dealing with the change of the argument of $f(z)$ along the curve $\gamma_z$ only i.e. $\Delta_{\gamma_{z}}\arg(f(z))=nk2\pi$! Or to wrap it up, you have 3 branch points $-1,1$ and due to the $\log$ also $\infty$! – Alexander Cska May 09 '20 at 07:35
  • You're not understanding here. There are branch points at $\pm1$ and at $\infty$. That is true. And if we cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$ we have $z+\sqrt{z^2-1}$ single valued and continuous on $\mathbb{C}\setminus $[-1,1]$. – Mark Viola May 09 '20 at 14:47
  • @MarkViola perhaps I am misunderstanding something. But just for clarity. $z+\sqrt{z^2-1}$ has no branch point at infinity (substitute $z=\frac{1}{t}$ and check $t \rightarrow 0$). The branch point at infinity comes from the $\log$ in the expression for $\arcsin$. However you can choose $(-\infty,-1)$ and $(1,\infty)$ because on the Riemann sphere there is always the possibility to connect two points by going vie the north pole. – Alexander Cska May 10 '20 at 14:07
  • @@alexandercska Do you really believe that $\sqrt {z^2}$ is single valued and has no branch point at $0$? – Mark Viola May 10 '20 at 15:11
  • @MarkViola I am not sure how exactly are you computing the branch points (would be interesting to learn), but to put it in this way $\sqrt{z^2-1}$ is a textbook example. I don't believe that the textbook is wrong :). Branch points are $-1,1$ how to connect them (via infinity or not) is matter of choice. – Alexander Cska May 10 '20 at 15:27
  • And $z=\infty$ IS also a branch point. As you said, substitute $z=1/t$ and observe the behavior of $\sqrt{\frac{1-t^2}{t^2}}$ near $t=0$. Clearly there IS a branch point at $t=0$ since the function $\sqrt{t^2}$ is multivalued (it can be $t$ or $-t$). Do you see now? – Mark Viola May 10 '20 at 15:34
  • @MarkViola well the argumentation that $t=0$ is wrong! No branch at infinity, unless you would like to jon -1,1 via the north pole of the Riemann sphere. The branches of $\sqrt{z^2-1}$ are in many textbooks, feel free to consult them. Have a nice day. I am don with this. – Alexander Cska May 10 '20 at 16:15
  • @MarkViola yes sure me and a bunch of textbooks are wrong :) Let's hope for a second opinion on this. – Alexander Cska May 10 '20 at 16:23
  • @MarkViola $\sqrt{z^2}$ is multi valued but has no branch points. btw. this is another standard textbook example. You can find the textbook explanation on this subject in the edit section of my original answer. – Alexander Cska May 10 '20 at 18:35
  • Let's try another way. I believe you'd agree that $\sqrt{z+1}$ has branch points at $-1$ and $\infty$, correct? And $\sqrt{z-1}$ has branch points at $1$ and $\infty$. So, the branch cuts for their, product $\sqrt{z+1}\sqrt{z-1}$ are formed from contours from $-1$ to $\infty$ and from $1$ to $\infty$. Their are an uncountable number that these cuts can be taken. See this answer for a more detailed discussion. So, $\sqrt{z^2-1}$ actually has two branch points at $\infty$,but they annihilate each other. – Mark Viola May 11 '20 at 16:05
  • Now that we've finished that "aside" discussion, let's go back to the original topic. We agree that $\log\left(z+\sqrt{z^2-1}\right)$ has branch points at $-1$ and $1$. And we agree, I believe, that $\sqrt{z^2-1}$ is holomorphic on $\mathbb{C}\setminus [-1,1]$, correct (This is "text book stuff,"right?)? Then, $z+\sqrt{z^2-1}$ is also holomorphic on $\mathbb{C}\setminus [-1,1]$. Correct so far? We also agree that $z+\sqrt{z^2-1}\ne 0$. So, how would one evaluate the inegral $$\oint_{|z|=R>1} \log\left(z+\sqrt{z^2-1}\right),dz$$ – Mark Viola May 11 '20 at 16:18
  • @MarkViola you made a simple claim, $\sqrt{z^2}$ has branch points. Well it does NOT have branch points and I told you why (see the edit of the answer). I think that you are trying to disprove the definition of the complex root and log function on a simply connected domain. I suggest you first understand why $\sqrt{z^2}$ has no branch points and then come again. – Alexander Cska May 11 '20 at 16:21
  • Well, it is a semantic. That is to say that $\sqrt{z^2}=\sqrt{z}\sqrt{z}$ has two branch points at $0$ (and two at $\infty$)), but they annihilate one another. But that was an aside discussion. Have a look at the two previous comments I posted and let me know your thoughts. – Mark Viola May 11 '20 at 16:25
  • @MarkViola for complex numbers $\sqrt{ab}$ is not necessarily $\sqrt{a}\cdot\sqrt{b}$. Another interesting textbook fact. – Alexander Cska May 11 '20 at 16:25
  • Actually, $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for any branch of $\sqrt{ab}$ and some branch of $\sqrt{a}$ and some branch of $\sqrt{b}$. That is they are equal in terms of set equivalence. But you know that, right? It is also "text book stuff." ;-) – Mark Viola May 11 '20 at 16:27
  • SeeThis answer, This answer, and This answer for primers. – Mark Viola May 11 '20 at 16:32
  • @MarkViola $\sqrt{z^2-1}$ has no branch point at infinity. However, as you might know on the extended complex plain, you can join -1,1 going via zero or via the north pole of the Riemann sphere. I tell you again understand $\sqrt{z^2}$ this answers all your questions – Alexander Cska May 11 '20 at 16:34
  • You wrote in your profile, "I really appreciate the possibility to discuss and get answers form the experts on this forum." Have you actually read the comments I've posted and the references I've provided with links? – Mark Viola May 11 '20 at 16:38
  • @MarkViola you clearly didn'd know that $\sqrt{z^2}$ has no branch points. Which drags all the rest. You keep bringing in some wrong argumentation and jumping from pint to point. I don't have the nerve to continue this "palaver". Wish you all the best! – Alexander Cska May 11 '20 at 16:42
  • To address your statement that $\sqrt{ab}\ne \sqrt{a}\sqrt{b}$, I have left you with three references that explain the idea of Set Equivalence. That is a "Text Book"concept. Do you really want to learn here? Evidently, you don't. And I have nothing to learn from you in this topic. – Mark Viola May 11 '20 at 16:43
  • @MarkViola ok expert appreciate it! I see where this is heading $\sqrt{z^2}$ :) – Alexander Cska May 11 '20 at 16:43
  • $\sqrt{z^2}$ has two branch points that annihilate one another; it is a semantic. That is why there is no requisite branch cut. And you don't know what you're talking about, do you? SET EQUIVALENCE!! – Mark Viola May 11 '20 at 16:45
  • @MarkViola $\sqrt{z^2}$ has no branch points because, drums ..... , the definition of the power function on a simply connected domain! The set equivalence is in the edit. More Palaver to come? – Alexander Cska May 11 '20 at 16:51
  • You are distracting from your unwillingness (or inability) to address in good faith the original two comments I posted today. – Mark Viola May 11 '20 at 16:52
  • There is no understanding you nor your cookbook approach to mathematics. I think that we need the comment of a real mathematician here. I am lost sorry. In such situations I say "Alter Schwede!". – Alexander Cska May 11 '20 at 16:56
  • When corona is over I will invite you to have some beers! – Alexander Cska May 11 '20 at 16:58
  • Well, that is a nice gesture. But I really wish that we could come together here. – Mark Viola May 11 '20 at 16:59
  • I wish I could join you. Do you live in the US? The Barolo sounds great! – Mark Viola May 11 '20 at 17:02
  • Thank god no :) – Alexander Cska May 11 '20 at 17:04
  • Well, despite our not coming together on this topic (yet?), I will consider you a friend on this site. If you go to my profile, I have provided a way to contact me outside this site if you like. – Mark Viola May 11 '20 at 17:07
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No, branch points and poles are different. A pole is an isolated singularity, a branch point is not (e.g. there is no way to define $\arcsin(z)$ as an analytic function in a punctured disk $\{z: 0 < |z - 1| < \epsilon\}$).

Robert Israel
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  • Sir, i was thinking of finding branch point of $\arcsin$ through the poles its derivative $\displaystyle\frac{1}{\sqrt{1-x^2}}$ ? since by Cauchy's theorem, the function will have different values when going around arbitrary small closed loop around such point ? – Geralt of Rivia Nov 21 '12 at 07:30
  • You can look at the series expansions of $f'(x) = 1/\sqrt{1-x^2}$ around each of its singularities. Around $x = 1$, for example, the expansion of $f'(x)$ (and therefore also of $f(x)$) involves half-integer powers of $x-1$. Around $x=\infty$, $f'(x) = \pm (i/x + i/(2x^3) + \ldots)$, and the term in $1/x$ gives $f(x)$ a logarithmic term. – Robert Israel Nov 21 '12 at 21:23
  • Hi Robert. I hope that you are staying safe and healthy. Just curious … How would one evaluate the integral $\oint_{C}\log\left(iz+\sqrt{1-z^2}\right),dz$ where $C$ is any rectifiable curve that encloses (one time) the branch points at $z=-1$ and $z=1$? It seems that $\arg\left(z+\sqrt{z^2-1}\right)$ and $\arg(z)$ lie in the same quadrant when we cut the plane from $-1$ to $\infty$ and from $1$ to $\infty$. Then, on $|z|=r>1$, as $\arg(z)$ goes from $0$ to $2\pi$ the argument of $z+\sqrt{z^2-1}$ does likewise. This would imply that there is a branch point of the logarithm ($z=0$). – Mark Viola May 06 '20 at 15:47
  • @RobertIsrael Hi Robert. I hope that you and your family are staying safe and healthy. I left a comment last week and would really appreciate hearing your thoughts. – Mark Viola May 11 '20 at 16:28
  • $\log(iz+\sqrt{1-z^2})$ has a branch point at $\infty$, so using any branch, your curve will cross at least one branch cut. The value of the integral will depend on where that occurs. – Robert Israel May 11 '20 at 17:34
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I believe we can think about this problem from the perspective of the domain, which is much easier. The expression $\sqrt{1-z^2}$ in $\sin^{-1}(z) = -i\log[iz + \sqrt{1-z^2}]$ is not explicit as the input is a complex number. The true definition is $\sqrt{1-z^2} = \exp[0.5 \log(1-z^2)]$, which directly implies that $1-z^2 \neq x$ for some real number $x \leq 0$ as the real part of the output of the exponential function must be strictly positive. Therefore, we have $z \notin [-\infty,-1] \cup [+1,+\infty]$, which is the branch cut of the arcsin function. In fact, all trigonometric functions possess similar branch cut due to the domain of the log function.

Dreamsky Van
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