6
$\begingroup$

Although our definition requires all analytic functions to be single-valued, it is possible to consider such multiple-valued functions as $\sqrt{z}$, $\log z$, or $\arccos z$, provided that they are restricted to a definite region in which it is possible to select a single-valued and analytic branch of the function.

For instance, we may choose for $\Omega$ the complement of the negative real axis $z\le 0$; this set is indeed open and connected. In $\Omega$ one and only one of the values of $\sqrt{z}$ has a positive real part. With this choice $w=\sqrt{z}$ becomes a single-valued function in $\Omega$; let us prove that it is continuous.

(The set $\Omega$ is the open set on which $f$ is defined.)

I don't really understand what all this means. Why is the negative real axis described by $z\le 0$ (shouldn't it be $x\le 0$?) Isn't it always the case that one value of $\sqrt{z}$ has a positive real part, since the two values are negative of each other? And why does $w=\sqrt{z}$ become a single-valued function, when we restrict the domain but not the range?

$\endgroup$
2
  • 1
    $\begingroup$ We do restrict the range here. But "slicing the complex plane along the negative real axis" is needed to allow us to restrict the range in such a way that we get a continuous function. $\endgroup$
    – Jyrki Lahtonen
    Aug 29, 2013 at 14:47
  • 3
    $\begingroup$ $z \leqslant 0$ is another way to express $\Re z \leqslant 0 \land \Im z = 0$. $x \leqslant 0$ would be $\Re z \leqslant 0$, which is the closed left half plane. Except for $z \leqslant 0$, one of the two square roots of $z$ has positive real part, and the other has negative real part. The point is that selecting the one with positive real part gives a continuous function on $\Omega$ (selecting the other one would too). You can't select one of the two square roots in such a way that you get a continuous function on all of $\mathbb{C} (\setminus \{0\})$. $\endgroup$
    – Daniel Fischer
    Aug 29, 2013 at 14:48

1 Answer 1

Reset to default
3
$\begingroup$

Why is the negative real axis described by $z\le 0$ (shouldn't it be $x\le 0$?)

Whenever a complex number appears in an inequality, the inequality implicitly says that the number is real. Hence, $z\le 0$ is indeed the negative real axis. The inequality $x\le 0$, on the other hand, would be understood as describing the left half-plane (since it does not say anything about $y$).

Isn't it always the case that one value of $\sqrt{z}$ has a positive real part

True (if "positive" means $\ge 0$). But after removing the negative semiaxes we can say the same, but with "positive" being $>0$. The difference is substantial. Strict inequalities are stable under small perturbations; thus, choosing the root with positive real part gives us a continuous function.

And why does $w=\sqrt{z}$ become a single-valued function, when we restrict the domain but not the range?

The range is determined by the domain. Whenever we talk about the restriction of some function, it's the domain that is being restricted.

There are still two values of square root to choose from, but it's now possible to make the choice continuously, (and, consequently, in a holomorphic way).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .