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I tried to find principal value for $\sqrt[3]{z}$ , I started from $$ z=w^3 $$ So $$ w_1=\sqrt[3]{r} \exp\left(\frac{Arg(z)}{3} i\right)$$ $$ w_2=\sqrt[3]{r} \exp\left(\frac{Arg(z)+2\pi}{3} i\right)$$ $$ w_3=\sqrt[3]{r} \exp\left(\frac{Arg(z)-2\pi}{3}i \right)$$ now which solution is a principal value?

I thought its $w_1$ which mean $Arg(\sqrt[3]{z})=\frac{Arg(z)}{3}$

but Wolfram-alpha says that $\sqrt[3]{-1-i}=\sqrt[6]{2}\exp\left(\frac{5\pi}{12}i \right)$ but $Arg(-1-i)=-\frac{3\pi}{4}$

So it must be as I think $Arg(\sqrt[3]{-1-i})=-\frac{\pi}{4}$ therefore $\sqrt[3]{-1-i}=\sqrt[6]{2}\exp\left(-\frac{\pi}{4}i \right)$

what is the wrong here ? is wolfram made wrong ?

RobPratt
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Faoler
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    What is your definition of a principal value ? – François Gatine Jul 14 '24 at 21:34
  • @FançoisGatine a principal value mean choose one of many possible values since the function must has one value for all domain

    in roots function we have in square root to solutions and the principal value has positive real part because of principal of $Arg(z)$ function

    in cubic roots from $r=\sqrt{x^2+y^2}$ we can get that principal value has positive real part

    and Im searching for better condition that is true for choosing a value that remains true to the definition $arg(z)$ and square root

     – Faoler Jul 14 '24 at 22:15
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    You can't make a continuous function $f$ over $\mathbb C$ such that $\forall x\in\mathbb C(f(x)^3=x)$. – Lucenaposition Jul 14 '24 at 23:19
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    Wolfram-Alpha will report the one you liked if the input is rewritten as (-1-i)^(1/3) . – Zack Fisher Jul 15 '24 at 00:15
  • To follow up on the comment of @FançoisGatine, your response to that comment does not give a well-defined description of the principle value. – Lee Mosher Jul 15 '24 at 22:57
  • @LeeMosher ok can you help me in that ? there must be only one value for cubic root So how can I define it and how to find ? – Faoler Jul 16 '24 at 10:14
  • Well, right there is the big problem. The thing to face when writing "There must be only one value for cubic roots" is a simple mathematical fact that you already know: the equation $z^3=a$ has three different complex valued solutions for any complex number $a \ne 0$. If you like, you can give a rule which selects one of those solution for each $a$. But just giving a name for a rule, like "principal value", is not sufficient if you don't also give the mathematical definition. – Lee Mosher Jul 16 '24 at 14:23
  • @LeeMosher choosing one of three solution is not randomly it must be follow what $Arg(a)$ and $\sqrt{a}$ defined So I used the name for that rule to show that is related for functions we used to calculate cubic root for a number> – Faoler Jul 17 '24 at 07:05
  • Fair enough, except for one point. Your title question is How to find principal value of the cube root, and your post itself asks "Is Wolfram wrong?" Each of those questions presumes that a definition for the principal value of the cube root already exists and is already known. – Lee Mosher Jul 17 '24 at 13:26
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    Closely related question: https://math.stackexchange.com/questions/322481/principal-nth-root-of-a-complex-number – Lee Mosher Jul 17 '24 at 13:26
  • @LeeMosher So you mean there is no Agreed definition for a principal value of cubic root ? and I checked your link and found "There really is not a coherent notion of "principal" nth root of a complex number" – Faoler Jul 17 '24 at 16:09
  • That's correct. The one described in the answer of @ZackFisher is a perfectly reasonable choice. One can make other choices. There is no particular mathematical reason to prefer one over the other, besides convention and/or agreement. And there are reasons not to insist on any one choice, as that answer in that link explains. – Lee Mosher Jul 17 '24 at 18:33
  • @LeeMosher So can be $\sqrt{1}=-1$ correct ? since there is no particular mathematical reason to choose $1$ or $-1$ . – Faoler Jul 17 '24 at 18:51
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    Interestingly, for the square root function defined on $[0,\infty)$, the mathematical community has long ago rallied around a convention, namely that if $x \ge 0$ then $\sqrt{x} \ge 0$. But for the square root function defined on $\mathbb C$, yes, I can imagine a convention producing the output $\sqrt{1}=-1$. For instance, one might wish to invert function $w=z^2$ with domain restricted to the left half plane $\text{Re}(z) < 0$. – Lee Mosher Jul 17 '24 at 19:00

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According to a comment on the similar question, the first non-real root as you traverse counter clockwise from the positive real axis is chosen by Mathematica.

Zack Fisher
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