help with $\;(x+3)^{1/3} = (x-1)^{1/2}$

Question

Taking the 6th power of both sides leads to
$x^2 + 6x + 9 = x^3 - 3x^2 + 3x -1 \;\Rightarrow\; 0 = (x-5)(x^2 + x + 2).$
Therefore, the only possible solutions are $\;\{5, \frac{-1}{2} \pm i\frac{\sqrt{7}}{2} \}.$

$x=5\;$ can be manually verified as satisfying the equation in this posting's title, but what is the easiest way to determine whether either or both of the other two candidate solutions also satisfy the equation?

Can this be done without having to solve a cubic equation to determine (for example) all 3 cube roots of $\; \left(\frac{5}{2} + i\frac{\sqrt{7}}{2}\right)?$

Furthermore, assume that all 3 cube roots of $\;\left(\frac{5}{2} + i\frac{\sqrt{7}}{2}\right)\;$ are identified and that both square roots of $\;\left(\frac{-3}{2} + i\frac{\sqrt{7}}{2}\right)\;$ are identified. In order to conclude that the equation in the title is satisfied, which of the 3 cube roots are you supposed to compare to which of the two square roots? Re-wording this last question, when $\;(x+3)\;$ and/or $\;(x-1)\;$ are complex, with at least one of the complex values having distinct roots, does this automatically render the equation ambiguous?

My Answer:

The interface won't allow me to post an answer, so I'm placing my conclusion here.

Construing $\;e^{i\theta}$ as $\cos(\theta) + i\sin(\theta),\;$ and using de Moivre's theorem, which is easily proven by induction, $\;\left(e^{i\theta}\right)^n = e^{n\times i\theta}.\;$

Therefore, for $\;n\in\mathbb{Z+},\;$ the equation $\;z^n = 1\;$ will have the n distinct roots $\xi_k = e^{i\times 2k\pi/n} \;: \;k \in \{0,1,\cdots,n-1\}.$

Further, any fixed non-zero complex $z_1$ can be expressed in the form $\;re^{i\theta},\;$ where $\;0<r\in\mathbb{R},\;$ and $\;\theta\in[0,2\pi).$

Thus, the equation $\;z^n - z_1 = 0,\;$ which can have at most n roots will have the root $\;z_2 = r^{1/n}e^{i(\theta/n)}.\;$ Therefore the equation $\;z^n - z_1 = 0,\;$ will always have exactly n distinct roots, given by $\;z_2\xi_k, \;: \;k \in \{0,1,\cdots,n-1\}.$

In real analysis, $5$ is considered to be an answer to the equation $\;(x+3)^{1/3} = (x-1)^{1/2}\;$ However, in complex analysis, 8 has 3 cube roots, while 4 has 2 square roots, and exactly one of the 3 cube roots coincides with one of the two square roots.

Therefore, in complex analysis, it seems reasonable to construe
$\;[f(z)]^{1/n} = [g(z)]^{1/m}\;:$ $\;f(z),g(z)$ are two polynomials with integer coefficients and $\;n,m \in \mathbb{Z^+}\;$
as intending that $z_1$ is a solution if and only if at least one of the $n^{th}$ roots of $f(z_1)$ coincides with one of the $m^{th}$ roots of $g(z_1).$

Let $\;z_1 = \frac{-1}{2} + i\frac{\sqrt{7}}{2}.\;$ From the analysis shown at the start of this question, $\;(z_1 + 3)^2 = (z_1 - 1)^3.\;$ Therefore, the six roots of $\;\{(z_1 + 3)^2\}^{1/6}\;$ exactly coincide with the six roots of $\;\{(z_1 - 1)^3\}^{1/6},\;$ and the arguments (i.e. angles) of these 6 roots will differ with each other by $(2\pi/6).\;$ Further, 3 of these 6 roots correspond to the 3 roots of $\;(z_1 + 3)^{1/3}\;$ and these 3 roots will have arguments that differ with each other by $(2\pi/3).\;$

Similarly, 2 of these 6 roots correspond to the 2 roots of $\;(z_1 - 1)^{1/2}\;$ and these 2 roots have arguments that differ with each other by $(2\pi/2).\;$

Therefore, one of the cube roots of $\;(z_1 + 3)\;$ must coincide with one of the square roots of $\;(z_1 - 1).\;$

Answer 1

You want to find all $x\in{\mathbb C}$ such that the set $$(x+3)^{1/3}\cap (x-1)^{1/2}\subset{\mathbb C}$$ is nonempty. In other words, you want to find all $x\in{\mathbb C}$ such that there is an $u\in{\mathbb C}$ with $$u^3=x+3,\qquad u^2=x-1\ .\tag{1}$$ You already have established that necessarily $$x\in\left\{5,-{1\over2}+i{\sqrt{7}\over2}, -{1\over2}-i{\sqrt{7}\over2}\right\}\ .\tag{2}$$ Now $(1)$ implies that, given $x$, the value of $u$ is determined by $$u={u^3\over u^2}={x+3\over x-1}\ .\tag{3}$$ The procedure is then clear: For each $x$ in the candidate list $(2)$ compute the corresponding $u$ by means of $(3)$; then check whether $u^2=x-1$. (This will automatically guarantee $u^3=x+3$ as well.) The candidates $x$ that pass this test can be considered as solutions to the original equation under the chosen interpretation.

Answer 2

Clearly, $5$ is a solution.

$\left|\frac{5}{2} + i\frac{\sqrt 7}{2}\right|^{\frac 13}$ and $\left|\frac{-3}{2} + i\frac{\sqrt 7}{2}\right|^{\frac 12}$ are respectively $(\sqrt 8)^{\frac 13}$ and $2^\frac 12,$ so they are equal.
Thus the three cube roots of $\;\color{red}{a=\frac{5}{2} + i\frac{\sqrt 7}{2}}\;$ and the two square roots of $\;\color{blue}{b=\frac{-3}{2} + i\frac{\sqrt 7}{2}}\;$ are located on the same $\color{violet}{\text{circle}}$ centered in $O,$ with radius $\sqrt 2.$

Due to the relative position of cube/square roots of a complex number, there can be at most one common root. The picture [drawn with GeoGebra] whispers $\color{violet}{w=-\frac 12 - i \frac{\sqrt 7}{2}}.$
Since $w$ verifies $$w^2=\frac{-3}{2} + i\frac{\sqrt 7}{2}\quad \text{and} \quad w^3=\frac{5}{2} + i\frac{\sqrt 7}{2},$$ it is solution of the given equation.

EDIT
One can proceed similarly with $\color{red}{\overline{a}}$ and $\color{blue}{\overline{b}}$ and conclude that $\color{violet}{\overline{w}}$ is a solution as well.

Conclusion
The set of solutions is $\;\{5, -\frac{1}{2} \pm i\frac{\sqrt{7}}{2}$}.

Answer 3

The answer depends in part on the meaning you give to fractional powers of complex numbers. If you think of $z^{1/n}$ as any complex number whose $n$th power is $z$, then you can get $(x+3)^{1/3}$ to equal $(x-1)^{1/2}$ with $x=(-1\pm i\sqrt7)/2$. But if you want $z^{1/n}$ to be a single-valued function, then the answer will depend on the convention you use for picking the $n$th root.

In particular, most $n$th-root conventions map the upper half plane into the first quadrant (if $n\gt1$). Under any such convention, the argument of $\left({5\over2}+i{\sqrt7\over2} \right)^{1/3}$ is less than $\pi/6$, while the argument of $\left({-3\over2}+i{\sqrt7\over2} \right)^{1/2}$ is greater than $\pi/4$, hence we do not have $(x+3)^{1/3}=(x-1)^{1/2}$ for $x={-1\over2}+i{\sqrt7\over2}$ for those conventions.