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This is incredibly basic, yet I cannot for the life of me confirm it. Let's say in the topic of complex numbers we are told to calculate the roots of:

$ z^4 = \sqrt 9 $

Are we to assume $ z^4 = 3 $ and calculate 4 roots or should we proceed with " $ z^8 = 9 $ ", and calculate 8 roots ? Since we are not operating in real numbers, the fact that a negative square root can exist in this instance confused me. Do we just take it as +3 since the square root function is specific to non-negative real numbers? I know the professor wanted us to calculate 4 roots, but I would like to make sure.

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    $\begingroup$ Honestly this depends a lot on how the square root symbol is defined in your context. Different mathematicians have different usage, so you’d better ask your professor. $\endgroup$ Commented Mar 9, 2021 at 6:39
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    $\begingroup$ As long as the square-root is treated as a real-valued operation that only produces a non-negative value, there is no issue, although we would usually just write the equation as $ \ z^4 \ = \ 3 \ $ ; there are only four roots of this equation. (Using $ \ z^8 \ = \ 9 \ $ will introduce spurious solutions to the original equation.) The radicals should really be avoided when we are working with complex numbers, as they tend to induce just the sort of confusion you're having (and confusion that is the source of a number of posts here). $\endgroup$ Commented Mar 9, 2021 at 6:44
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    $\begingroup$ It might be worth explicitly stating that you have followed the convention that $\sqrt{9}$ takes only the positive value and then find the 4 values of $z$. $\endgroup$ Commented Mar 9, 2021 at 6:52
  • $\begingroup$ Convention has it that if $w \in \mathbb R^+$ then $\sqrt w$ is the principal root, whereas if $w\not \in \mathbb R^+$ then $\sqrt w = \{z|z^2 =w\}$ a set of two elements. So by that convention $z^4 = \sqrt 9 =3$ has four roots. BUT how did the question come about. If you had $z^4 = \sqrt{\alpha};\alpha \in \mathbb C$ you have $8$ roots. But then if we did further calculations and found $\alpha =9$.... well the original intent of $\sqrt{alpha}$ was the set of all roots so the solution should be $z^4 =\pm{\sqrt 9}$ has eight roots. $\endgroup$ Commented Mar 9, 2021 at 8:23
  • $\begingroup$ "The radicals should really be avoided when we are working with complex numbers, as they tend to induce just the sort of confusion you're having" But the problem is what to replace them with....sigh.... $\endgroup$ Commented Mar 9, 2021 at 8:28

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In light of the comment that the original expression was $z^4 = \sqrt{\frac {-9}{i^2}}$ that... this is almost more a case for a lawyer than a mathematician.

The convention in complex analysis is that $\sqrt[p]{\alpha}:= \{z\in \mathbb C| z^p = \alpha\}$ with a practical (yet arbitrary) exception that if $\alpha \in \mathbb R^+$ then it does not mean the set of all roots be merely $\sqrt[p]{\alpha} = $ the single element $w\in \mathbb R^+$ so that $w^p=\alpha$.

So $z^4 = \sqrt{9}\implies z^4 =3\implies$ 4 complex roots.

!BUT! now the lawyer: If $z^4 = \sqrt{\frac {-9}{i^2}}$ it is not declared and (for some values of "clear") not immediately clear then $\frac {-9}{i^2} \in \mathbb R^+$ (it is, but it's not immediately apparent). As the exception for $\alpha \in \mathbb R^+$ was a practical one to mesh with the conventions of "regular" non-complex math, the expression $\frac {-9}{i^2}$ clearly indicates that we are doing with the full range of complex numbers.

So I'd interpret $\sqrt{\frac {-9}{i^2}} = \{z|z^2 =\frac {-9}{i^2}\}=\{3,-3\} \ne 3 = \sqrt 9$. And that $z^4 =\sqrt{\frac {-9}{i^2}}\implies z^4 \in \{3,-3\}\implies $ 8 complex roots.

And we'd have a really weird case where $\alpha = w$ but $\sqrt{\alpha} \ne \sqrt w$ which... well, that's a deliberate manipulation of contextual meaning taken deliberately out of context. It doesn't count.

Not all books will agree with my assessment. Ask your professor.

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I think most text books consider that if $w> 0$ (which implies $w \in \mathbb R$) that $\sqrt[k]{w}$ or $w^{\frac pq}$ refers to the principal positive root, While for any other case, $\sqrt[k]{w}$ or $w^{\frac pq}$ would refer to the set of all roots.

Thus that would mean $\sqrt 9 = 3$. And you are being asked nothing more than to solve $z^4 = 3$>

But... you should ask your professor.

After all, if it is consistent for if $w \not \in \mathbb R^+$ to have $\sqrt[k]{w}$ or $w^{\frac pq}$ be that multi-valued set of roots, a text could argue for consistency it would mean that for $w \in \mathbb R^+$ as well.

If so the $\sqrt{9} = \{3, -3\}$ and you are being asked to solve ($z^4 =3$ OR $z^4 = -3$) which, as you point out would be the same as solving $z^8 = 9$.

.....

So which does your text and professor do?

I'd say in about 80% of the cases, it would mean $\sqrt{9} = 3$ as $9\in \mathbb R^+$; a conventional inconsistency.

.....

But if so... why is this question even being asked? Why not just say $z^4 = 3$. Is this part of something more complicated?

If you had $z^4 = \sqrt{\alpha};\alpha \in \mathbb C$ you have $8$ roots. But then if we did further calculations and found $\alpha =9$.... well the original intent of $\sqrt{alpha}$ was the set of all roots, and suddenly stating $z^4 = \sqrt 9$ is a bait and switch and the meaning has changed. We can't just change definitions mid calculation just because our number falls into an inconsistant and arbitrary notational convention. so the solution should be $z^4 =\pm{\sqrt 9}$ has eight roots.

.......

On third reading of Levinson/Reheffer "Complex Variables" They state the $z$ so that $z^q = \alpha^p$ are called the $q$th roots all of these is designated by $\sqrt[q]{\alpha^p}$ or by $\alpha^{\frac pq}$. They then state that if $\alpha > 0$ (I don't know why they didn't just say if $\alpha$ is positive real) then $\alpha^{\frac pq}$ is reserved for the one with $\arg$ of $0$ (why the didn't just say the positive real one I don't know) unless stated otherwise.

I see that they did not make the same claim for the notation $\sqrt[q]{\alpha^p}$. I do not know if that was an oversight or if the specifically mean that the $\sqrt{}$ symbol is not reserved for the positive value.

That'd mean $\sqrt{9} = \pm 3$ and $8$ roots.

I may have been inaccurate in my assessment that 80% of text would believe $\sqrt 9 = 3$. It could be that they interpret $9^{\frac 12} =3$ but $\sqrt{9} = \{3,-3\}$.

After we kill all the lawyers we should kill all the mathematicians.

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  • $\begingroup$ Thank you for the detailed answer. To clarify, the original question is not much more complicated it asks us to calculate the roots of: $ z^4 = \sqrt{\frac{-9}{i^2}} $ $\endgroup$ Commented Mar 9, 2021 at 10:18
  • $\begingroup$ Oh... boy.... I'd say that as in the original expression it is not given that $\frac {-9}{i^2}$ will be found to be a positive real number (although it obviously will) that $\sqrt{\frac{-9}{i^2}}$ is by definition $\{w\in \mathbb C| w^2 = \frac {-9}{i^2}\}=\{w\in\mathbb C|w^2=9\}=\{3,-3\}=\pm 3$ and that there will be $8$ solutions. $\endgroup$ Commented Mar 9, 2021 at 15:42
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$z^4=\sqrt9$ which implies: $z^4=3$ and $z^4=-3$

Solving: $z^4=3$

Let $u=z^2$, then $u^{2}=3 \rightarrow u=\pm \sqrt 3$

Now, $z^2=\sqrt3$ and $z^2=-\sqrt3$. Thus, there are four roots $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3}$$

Solving $z^4=-3$

Let $u=z^2$, thus $u^{2}=-3 \rightarrow u=\sqrt3i \,\,\,\&\,\,\, u=-\sqrt3 i$

Hence, $z^2=\sqrt3i$ and $z^2=-\sqrt3 i$

Let $z=x+iy$, for $x,y \in \mathbb{R}$. Then $z^2=\left(x^2-y^2\right)+i(2xy)$.

Case 1: $z^2=\sqrt 3 i$

Now, the real part equals the real part, and the imaginary part equals the imaginary part. Hence, $$x^2-y^2=0\\ x=\pm y\\x=y$$ and $$2xy=\sqrt 3 \\2x^2=\sqrt 3 \\x=\pm\sqrt\frac{\sqrt 3}{2}\\ x=\pm\frac{\sqrt[4]{3}\sqrt{2}}{2}$$ and plugging back in these $x$-values into $2xy=\sqrt3$, we find that for $x=\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=-\sqrt[4]{\frac{3}{4}}$ and for $x=-\frac{\sqrt[4]{3}\sqrt{2}}{2}$, $y=\sqrt[4]{\frac{3}{4}}$. Hence writing them in terms of $z$, we get: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$

Case 2:

$z^2=-\sqrt 3 i$. Similarly, for $x,y \in \mathbb{R}$ $x^2-y^2=0\\ x=\pm y\\ x=-y$ and $2xy=-\sqrt 3\\-2x^2=-\sqrt 3\\ \therefore x= \pm \frac{\sqrt[4]{3}\sqrt{2}}{2}$ corresponding to $y=\pm \sqrt[4]{\frac{3}{4}}$. Hence the solution in terms of $z$ are: $$z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i$$

Overall, there are $8$ roots to this equation: $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3} \,\,\,\&\,\,\, z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i \,\,\,\&\,\,\,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i $$

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A definition of the square root $x$ of a given $y$ is $x$ such that $x\cdot x=y$. This definition works over any field.

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  • $\begingroup$ Yes, but what's the definition of the symbol $\sqrt{}$ when what is under it is a positive real number? Is the convention the symbol refers to the set of all square roots or does the symbol only refer to one particular (the positive one) square root. $\endgroup$ Commented Mar 9, 2021 at 15:38
  • $\begingroup$ $\sqrt{x}$ only refers to the positive square root (hence $\pm\sqrt{b^2-4ac}$ in the quadratic formula). @fleablood $\endgroup$ Commented Mar 9, 2021 at 15:51
  • $\begingroup$ But then what is the definition of $\sqrt{1 + 2i}$. That's the issue. If $z^4 = \sqrt{1+2i}\iff z \in \{w|w^4\in \{v|v^2=1+2i\}\}=\{w|w^8=1+2i\} \implies z^8=1+2i$ then is it consistent to say that $z^4 = \sqrt 9\implies z^4 = 3$. We are using $\sqrt{}$ differently in those calculations. Is that kosher? $\endgroup$ Commented Mar 9, 2021 at 16:11
  • $\begingroup$ "A definition of the square root x of a given y is x such that x⋅x=y" But there is going to be two such $x$. Which one is the square root. $\endgroup$ Commented Mar 9, 2021 at 16:12
  • $\begingroup$ @fleablood; complex numbers don't possess an order, so 'positive' doesn't mean anything here. There are two square roots (due to the fundamental theorem of algebra), each one is as good as the other. $\endgroup$ Commented Mar 9, 2021 at 16:24

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