On the real square root and branches of the complex square root.

When we think of the real square root we think of it as multivalued. The real(and complex) square root of $4$ is $\pm 2$.

But when defining the branches of the square root in $\mathbb{C}$ we usually remove the negative axis. This look couterproductive since the most basic extension that we want i.e the negative reals vanishes which are the once we think of when defining $i$.

Are these branches linked somehow or are they seperate things?

2 Answers

If we want to deal with the square root as a function, we must make a choice concerning which square root we have in mind. If we are deling with non-negative real numbers, thare is a natural choice: we take the non-negative square root.

If we are dealing with complex numbers, then there is no natural choice; it depends upon the context. You wrote that “we usually remove the negative axis”. Indeed, but there are infinitely many choices. Suppose, for intance, that we remove the numbers of the form $ai$, with $a\leqslant0$. Then every remaining complex number $z$ can be written as $r\bigl(\cos(\theta)+i\sin(\theta)\bigr)$, with $r>0$ and $\theta\in\left(-\frac\pi2,\frac{3\pi}2\right)$, and we can define $s(z)=\sqrt r\left(\cos\left(\frac\theta2\right)+i\sin\left(\frac\theta2\right)\right)$. This function $s$ is also a square root, and it is a continuous (and even an analytic) one.

In order to work with a single square root function, we must leave the complex complex plane and work in a Riemann surface instead.

• Does multivaluedness of the square root have anything to do with the solutions to $x^2-4=0$ which is mentioned above? To me it looks like we apply a function and to aplly it we have to define it right? – user517104 Dec 29 '17 at 10:39
• A part from that you say that we pick branch depending on context right?. It strikes me as kinda of stupid to have a convention in $\mathbb{C}$ which makes the situation in the real case undefined but I guess thats just the way things turned out. I would be nice if the convetion in $\mathbb{R}$ would be compatible with the one in $\mathbb{C}$ – user517104 Dec 29 '17 at 10:42
• @l33tg33k I dislike the word “multivalued”. In this context, it's just a complicated way of saying that every complex number (other than $0$) has two and only two square roots. – José Carlos Santos Dec 29 '17 at 10:44
• right, but it is werid that we have diffrent convetions right? – user517104 Dec 29 '17 at 11:00
• @l33tg33k Which different conventions are you talking about? – José Carlos Santos Dec 29 '17 at 11:12

The real square root of a nonnegative real number $a$ is usually defined as the nonnegative number $b$, which square is $a$. In your example it's $2$ and not $-2$. But the solutions of the equation $x^2-4=0$ are $\pm 2$. Do you see the difference between square root and solutions of an equation? It's not multivalued, but in $\mathbb{C}$ it is. Make a copy of $\mathbb{C}$ and remove, for example, the nonpositive real axis and link them together. You can see a picture of the square root branches here. They are linked together.

• I dont see the difference between the square root and the solution to that equation. Its solution is $x=\sqrt{4}$ – user517104 Dec 29 '17 at 10:19
• The square root of a nonnegative real number is only one number, it's unique! The solutions of the corresponding equation are more than one number. Example: $\sqrt{4}=2$ but $x^2=4$ has the solutions $\pm \sqrt{4}=\pm 2$ – Fakemistake Dec 29 '17 at 10:21
• So the multivaluedness in $\mathbb{C}$ has nothing to do with the solutions of the equation? – user517104 Dec 29 '17 at 10:24
• if we remove or solve the equation we apply the "function" $f(x)=x^{\frac{1}{2}}$ right? – user517104 Dec 29 '17 at 10:32
• @l33tg33k We apply the functions $f(x)=\sqrt{x}$ and $f(x)=-\sqrt{x}$. These are the two branches of the square root function. – Fakemistake Dec 30 '17 at 8:12