Roots of the equation $|x|^{n-2}x^2 +\lambda = 0$, $\lambda\in(0,\infty)$ I want to find all the roots of the equation $$|x|^{n-2}x^2  +\lambda = 0,$$ where $\lambda\in (0,\infty)$.

I'd like to know how I can find them.
Note When this answer was written, the OP's question was to solve the equation where $\lambda\equiv 1$. The OP since changed the question after this answer was given.


Clearly $x\neq 0$ for otherwise $1=0$ which is absurd. Hence, let $x=re^{i\theta}$ where $r&gt;0$, then

$$|re^{i\theta}|^{n-2}r^2e^{2i\theta}+1=0,$$
giving
$$|r|^{n-2}|e^{i\theta}|^{n-2}r^2e^{2i\theta}+1=0.$$
Since $r&gt;0$ and $|e^{i\theta}|=1$, then
$$r^ne^{2i\theta}+1=0.$$
Hence,
$$r^n\cos(2\theta)+ir^n\sin(2\theta)+1=0.$$
We obtain the system of equations,
$$\left\{\begin{array}{rll}r^n\cos(2\theta)+1&amp;=&amp;0\\\sin(2\theta)&amp;=&amp;0\end{array}\right.$$

From the sine equation we see that $\theta=\pi k$ or $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$, in which case $\cos(2\theta)$ equals $1$ and $-1$ respectively.

Hence, when $\cos(2\theta)=1$, then $r^n=-1$, which is absurd since $r&gt;0$ and $n\in\mathbb{Z}$. When $\cos(2\theta)=-1$, then $r^n=1$ which implies $r=1$.

So, the only solutions are $r=1$ and $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$.

Therefore, $$x=re^{i\theta}=1\cdot e^{i\frac{\pi}{2}(2k+1)}=\pm i,$$
as you quite rightly pointed out.

UPDATE

The only solution for $\lambda=0$ is $x=0$. Suppose $\lambda&gt;0$, then we obtain the slightly different system of equations,
$$\left\{\begin{array}{rll}r^n\cos(2\theta)+\lambda&amp;=&amp;0\\\sin(2\theta)&amp;=&amp;0\end{array}\right.$$

From the sine equation we see that $\theta=\pi k$ or $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$, in which case $\cos(2\theta)$ equals $1$ and $-1$ respectively.

Hence, when $\cos(2\theta)=1$, then $r^n=-\lambda$, which is absurd since $r,\lambda&gt;0$ and $n\in\mathbb{Z}$. When $\cos(2\theta)=-1$, then $r^n=\lambda$ which implies $r=\lambda^{1/n}$.

So, the only solutions are $r=\lambda^{1/n}$ and $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$.

Therefore, $$x=re^{i\theta}=\lambda^{1/n}\cdot e^{i\frac{\pi}{2}(2k+1)}=\pm \lambda^{1/n}i.$$
Note the following $$x^2 = |x|^2 $$ Therefore the equation becomes $|x|^n=-1$. 

$-1 = e^{(2k-1)i\pi}\hspace{1cm}\text{ for k in {1,2..n}}$ 

So the roots are $$|x| = e^{\frac{(2k-1)i\pi}{n}}\text{for k in {1,2,..n}}$$ or $$x = \pm e^{\frac{(2k-1)i\pi}{n}}\text{for k in {1,2,..n}}$$