Does the square root function behave the same when calculating complex roots? 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.
 A: 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.
A: $z^4=\sqrt9$ 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: $$z=\pm\sqrt[4]{3}i \,\,\,\&\,\,\, z=\pm\sqrt[4]{3} \,\,\,\&\,\,\, z=-\frac{\sqrt[4]{3}\sqrt{2}}{2}+\sqrt[4]{\frac{3}{4}}i \,\,\,\&\,\,\,\:z=\frac{\sqrt[4]{3}\sqrt{2}}{2}-\sqrt[4]{\frac{3}{4}}i $$
A: 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.
