I'm trying to grasp the difference between branches for the complex square root and I'm having difficulty with some very basic examples.

First example, if I choose $\sqrt{\,}$ to denote the branch defined on $(-\pi,\pi)$:
$a = e^{\frac{2}{5}\pi i}$ and $b = e^{\frac{2}{3}\pi i}$
I get that $a\sqrt{b} \ne \sqrt{a^2b}$, because:
$$a\sqrt{b} = e^{\frac{2}{5}\pi i} \cdot e^{\frac{1}{3}\pi i} = e^{\frac{11}{15}\pi i}$$

$$\sqrt{a^2b} = \sqrt{e^{\frac{4}{5}\pi i}\cdot e^{\frac{2}{3}\pi i}} = \sqrt{e^{\frac{22}{15}\pi i}} = \sqrt{e^{\frac{-8}{15}\pi i}} = e^{\frac{-4}{15}\pi i}$$

notice the step in the end of the last line where I need to fix the exponent by subtracting $2\pi$ so that the exponent will be between $-\pi$ and $\pi$. If I didn't "fix" it, I would've got the result I wanted.

Well this obviously didn't work. so my question is what failed and for which branches the common rules for powers work like for real-valued roots and powers?

Second example, which is related but a bit more complicated:
Calculate: $$ \int_{|z|=5} \frac{1}{\sqrt{z^2+11}} dz$$ where $\sqrt{\,}$ denotes the branch for which $\sqrt{36}=-6$.

Firstly, it is not obvious to me if $\sqrt{z^2+11}$ is even meromorphic, because $z^2+11$ has roots in the interior of the contour, so why is this integral defined correctly?

Secondly, the solution I saw was as follows:
$$ \int_{|z|=5} \frac{1}{\sqrt{z^2+11}} dz \overset{w=\frac{1}{z}}{=} \int_{|w|=\frac{1}{5}} \frac{1}{w^2\sqrt{\frac{1}{w^2}+11}} dw = \int_{|w|=\frac{1}{5}} \frac{1}{-w\sqrt{1+11w^2}} dw $$

Now I'm supposed to understand that $\sqrt{1+11w^2}$ is analytic in the interior of the contour, which is reasonable because $w$ is close to $0$, so $1+11w^2$ is close to 1 and we can remove a ray from the origin. Is this the right explanation?
After agreeing with the last claim, the integrand only has a simple pole at 0 and from the residue theorem the answer will be: $$ 2\pi i \cdot Res_0\left(\frac{1}{-w\sqrt{1+11w^2}}\right) = 2\pi i \cdot\frac{1}{-\sqrt{1+11\cdot0}} = 2\pi i \frac{1}{-\sqrt{1}} = 2\pi i$$

Another question about this example arises from the first example. why is it correct to move the $w$ inside the root like so (and multiplying by $-1$) and is it correct that if $\sqrt{a^2}=-a$ for a single real $a>0$, then the same rule applies to all complex numbers?

I'm missing a lot of formality in this subject and I'd like to understand the reasoning behind those certain steps and also how in general I could understand those definition or maybe translate them to a problem with branches of log, which I'm more comfortable with (although not so much).

  • 1
    $\begingroup$ With such a function $f$ with different branches, once you have fixed a branch around $z=5$, it is natural to interpret $\int_{|z|=5}f(z)dz$ as the integral of the analytic continuation of $f$ along $5 e^{it} ,t \in [0,2\pi]$, then the rules of change of variable hold in that setting and that $\sqrt{1+11 w^2}$ is analytic on $|w|=1/5$ implies $f$ was analytic on $|z|=5$ $\endgroup$
    – reuns
    Jul 28, 2019 at 21:01
  • $\begingroup$ @reuns what do you mean by "the analytic continuation of $f$ along $ 5e^{it}, t\in[0,2\pi]$"? it looks to me like you chose a branch different from the one given in the question, which is the branch such that $\sqrt{36}=-6$ (either way, I'm also not sure that a branch is completely defined by such an assumption) $\endgroup$
    – GuySa
    Jul 29, 2019 at 8:32
  • 1
    $\begingroup$ I mean the analytic continuation along a curve. upload.wikimedia.org/wikipedia/commons/thumb/b/b8/… $\sqrt{36}=-6$ defines uniquely a branch of the square root around $36$ but not a branch of your function on the circle $|z|=5$. $\endgroup$
    – reuns
    Jul 29, 2019 at 13:22
  • $\begingroup$ @reuns I haven't heard of this idea of analytic continuation along a curve, but thinking about it now makes me think that it is enough to note that $\sqrt{36}=-6$, because in $36$'s neighborhood this branch is chosen, and afterwards every circle that intersects $36$'s neighborhood will also have to be defined by the same branch, because they need to agree on their intersection, etc... and then iteratively we get that all of the curve has to be defined like so $\endgroup$
    – GuySa
    Jul 29, 2019 at 16:57

2 Answers 2


If you make the branch cut along $(-\infty,0]$, then we obtain two branches for the complex square root. They are both functions that can be defined on all of $\mathbb C$, but not as a continuous function when you move in $\mathbb C$ across the ray $(-\infty,0]$. Lets call them for $z=r e^{i\theta}$, $\theta \in [-\pi,\pi)$, $$\sqrt[(1)]{re^{i\theta}}:= \sqrt r e^{i\theta/2},\\ \sqrt[(2)]{re^{i\theta}}:= \sqrt r e^{i\theta/2+\pi i}=-\sqrt[(1)]{re^{i\theta}}. $$ I'm not sure that the sentence

choose $\sqrt{\ }$ to denote the branch defined on $(-\pi, \pi)$

chooses a branch, but I think you meant to say that you wanted to choose what I called $\sqrt[(1)]{\,}.$ What "went wrong" is that you "went across" the branch cut. If $\sqrt[(1)]{\,}$ was continuous on all of $\mathbb C$, then there would be no need for cutting out branches. If you want to go past a branch cut without an abrupt change in values, what you should do is switch to the other branch. This is your "without the fix" result- $$\sqrt[(2)]{a^2b} =a\sqrt[(1)]{b}.$$ This is exactly what happens when you analytically continue along a path as in reuns's comment, i.e. if $f(t) = \sqrt[(1)]{e^{it}}$ is defined for $t\in (-\pi,\pi)$, then its analytic continuation (also called $f$) has $$ f(\arg b) = \sqrt[(1)]{b} \text{ but}\\ f(2\arg a + \arg b) = \sqrt[(2)]{a^2b} =a\sqrt[(1)]{b}.$$

(strictly speaking you analytically continue to larger and larger open subsets of $\mathbb C$, so instead of $t\in (-\pi,\pi)$ take a small open set containing this interval.)

In the integral, you are asked to use $\sqrt[(2)]{\,}$. What is presumably happening (I haven't checked) when you multiply by $w^2$ inside $\sqrt[(2)]{\,}$ is that it always pushes you to the other branch, regardless of the value of $w$.

The sketch of reuns's comment - define $$f:(-\pi,-\pi/2)\to \mathbb C, \quad f(t) = \sqrt[(2)]{25e^{i2t} + 11}$$ we find that applying this formula blindly in a neighbourhood of $-\pi/2$ leads to a discontinuity at $t=-\pi/2$. Analytically continuing instead, we end up with $$ f:\mathbb R \to \mathbb C,\\ \quad f(t) = \begin{cases} \phantom{\Big(}\!\!\!-\sqrt[(2)]{25e^{i2t} + 11} & t \in (-\pi/2,\pi/2) + 2k\pi\\ \sqrt[(2)]{25e^{i2t} + 11} & t \in (-\pi/2,\pi/2) + 2(k+1)\pi \end{cases} $$ And you should be able to integrate $$ \int_{-\pi}^\pi f(t)5ie^{it} dt. $$

  • $\begingroup$ (Don't have time to finish the computation right now...) $\endgroup$ Jul 29, 2019 at 14:10
  • $\begingroup$ Thank you for answering my questions, but I'm having difficulty in calculating the integral with the $f$ you defined in the end. Are you saying the solution I gave originally is incorrect or are you saying what formally happened with the branches in the question? $\endgroup$
    – GuySa
    Jul 29, 2019 at 16:45
  • $\begingroup$ @GuySa more the second, I have not checked if the integral is actually right. I looked a little to find an analogous calculation to verify the minus sign appearing, the best I found was this math.stackexchange.com/questions/1560201/… though its not quite the same $\endgroup$ Jul 30, 2019 at 1:55

Regarding the first question, consider what happens for the principal branch. By definition, $$a \sqrt b = |a| \, \sqrt {\smash [b] {|b|}} \, e^{i (2 \arg a + \arg b)/2}, \\ \sqrt {a^2 b} = |a| \, \sqrt {\smash [b] {|b|}} \, e^{i \arg(a^2 b)/2}.$$ Now, $\arg(a^2 b) = 2 \arg a + \arg b + 2 \pi k$ for some $k \in \{-1, 0, 1\}$. If $k \neq 0$, dividing by $2$ will introduce a factor of $-1$. Thus $$a \sqrt b = \sqrt {a^2 b} \cases { 1 & $-\pi < 2 \arg a + \arg b \leq \pi$ \\ -1 & otherwise}.$$ The second question implicitly assumes that you need to choose a branch which is continuous on $|z| = 5$, otherwise the result is not unique. A simple way to construct such a branch is $$f(z) = -\frac 1 {z \sqrt {1 + \frac {11} {z^2}}},$$ where $\sqrt z$ again denotes the principal branch (as requested, $f(5) = -1/6$). The square root in the denominator is regular at infinity and tends to $1$, therefore $\operatorname{Res}_{z = \infty} f(z) = 1$ and the value of the integral, assuming ccw direction, is $-2 \pi i$ (since the solution that you refer to takes the residue at zero, therefore also assumes ccw direction of integration, $2 \pi i$ is not correct).


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