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In particular, is it OK to write $z^{1/2} = - z^{1/2}$, where $z \in C$?

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2 Answers2

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No, not at all. First of all, $z^{1/2}$ is ambiguous, since every non-zero complex number has two square roots. So, unless you explain first how you are choosing a square root, don't use $z^{1/2}$ and, for the very same reason, don't use $\sqrt z$.

Besides, in $\mathbb C$, $a=-a\iff a=0$.

José Carlos Santos
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  • Does the ambiguity of $z^{1/2}$ not make it okay to write $z^{1/2} = -z^{1/2}$ because $z^{1/2}$ represents two values, which are basically $\sqrt{z}$ and $-\sqrt{z}$? Similarly, $a = -a \implies 2a=0 \implies a=0$ for complex numbers, but can one actually say, $z^{1/2} + z^{1/2} = 2z^{1/2}$? – A B Dec 16 '19 at 12:05
  • No, it doesn't make it okay. If you want that $z^{1/2}$ is a number, then you simply cannot have it equal to its symmetric, unless that number is $0$. – José Carlos Santos Dec 16 '19 at 12:37
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Usually, when we write an algebraic expression such as $z^{1/2}$, we are assuming that it has a unqiue value. For example writing something like $z^{1/2}=i=-i$ will give you some weird looks, to say the least. Sure, what you mean by this could be that $i$ and $-i$ are the values which satisfy $z^2=-1$ (which is true), but this is not how you denote it. For the same reason $z^{1/2}=-z^{1/2}$ is not acceptable notation.

Furthermore, as José states, writing $z^{1/2}$ is already frowned upon as it is not clear which square root of $z$ is intended (since there are two possible values). So to even use this notation, be sure to define it first (e.g. as the root with the smallest argument).

YiFan Tey
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  • "Sure, what you mean by this could be that $i$ and $-i$ are the values which satisfy $z^2=-1$..." That is exactly what I meant when asking this question. Thanks for information about the general perception of using $z^{1/2}$. This was essentially a query about the acceptable notations. Having said that, is it possible to interpret this equation as basically stating, $z^{1/2} = \pm \sqrt{z} = -z^{1/2}$; and does it then become acceptable in this regard? Also, often in textbooks, especially Physics textbooks, $a^{1/2}$ is used instead of $\pm\sqrt{a}$; it is also incorrect notation? – A B Dec 16 '19 at 12:14
  • @AB Note that notation is correct if and only if it is generally "perceived" as being acceptable. So the "general perception" of $z^{1/2}=-z^{1/2}$ being frowned upon can be rephrased as saying that it is plain wrong. No, $z^{1/2}=\pm\sqrt{z}=-z^{1/2}$ is not acceptable. Although it is true that $\pm i$ are the roots of $z^2=-1$, this is not how you denote it (like I said in the part of that sentence you didn't quote)! And I strongly doubt any textbooks use $a^{1/2}$ in place of $\pm\sqrt{a}$, and strongly suspect they've properly defined $a^{1/2}$ to mean one value and one value only. – YiFan Tey Dec 16 '19 at 12:20
  • Thanks for clearing that up. In Introduction to Quantum Mechanics by Griffiths, there is an equation on page 23, which goes like: $\int_0^h \frac{1}{2 \sqrt{hx}} = \frac{1}{2 \sqrt{h}} (2 x^{1/2}) |_0^h$. Would it be more appropriate to write: $\int_0^h \frac{1}{2 \sqrt{hx}} = \frac{1}{2 \sqrt{h}} (2 \sqrt{x}) |_0^h$ ? – A B Dec 16 '19 at 12:31
  • @AB Is $x$ here a real or complex quantity? – YiFan Tey Dec 16 '19 at 12:36
  • A real quantity. – A B Dec 16 '19 at 12:37
  • @AB Indeed, so there is no ambiguity here whatsoever. The integral is taken over the interval $[0,h]$ (assuming $h>0$), and $x^{1/2}$ for real $x\geq0$ is understood to be the positive square root of $x$. So in this case $x^{1/2}$ and $\sqrt{x}$ are equivalent. – YiFan Tey Dec 16 '19 at 12:40