I have following ODE, with the constraint that $n\neq1$: $$\bigg(y'\bigg)^{n-1}=ny^{n-1}$$ This is easily solveable in $\mathbb{R}$, however in this case both $y(z)$ and $n$ can be complex.
I would really like to just take the (n-1)th root on both sides. However since taking the root of a complex numbers yields multiple correct answers, I'm worried that I will miss out on solutions for $y(z)$. Is there a correct way to proceed here?
$$(y’)^{n-1}=ny^{n-1} \iff\left(\frac{y’}y\right)^{n-1}=n$$
Remember the complex logarithm, $k\in\Bbb Z$:
$$(n-1)\ln\left(\frac{y’}y\right)=\ln(n)+2\pi i k\\\ln\left(\frac{y’}y\right)=\frac{\ln(n)}{n-1}+\frac{2\pi i k}{n-1}\\\frac{y’}y=\sqrt[n-1]n e^{\frac{2\pi i k}n}$$
Now integrate:
$$\int\frac{y’}y dz=\int\sqrt[n-1]n e^{\frac{2\pi i k}{n-1}}dz\\\ln(y)= \sqrt[n-1]n e^{\frac{2\pi i k}{n-1}}z +C$$
Finally, exponentiate and set $e^C=c$. Therefore:
$$y=ce^{\sqrt[n-1]n e^{\frac{2\pi i k}{n-1}}z},k\in\Bbb Z$$