Find all complex solutions $z^{10} + 2z^5 + 2 = 0$.

Question

So I’m working on this equation $z^{10} + 2z^5 + 2 = 0$ to find all complex solutions, and I think I managed to solve it, but I can’t find solution manual for it, since it is really old exam task. The thing that makes me uncomfortable with my solution is that, shouldn’t I get just 10 solutions? But when I put in all $k$ values($k = 0,1,2,3,4$), you get 12 different angle solutions. Isn’t that wrong?

My answer:

$$ \sqrt{\mathstrut 2}^{1/5}e^{\frac{\left(\pm\frac\pi4i+2\pi k \right)}5} $$

Sorry could't figure out how to put 5 in denominator of the polar formula...

Answer 1

solve for $$w^2 + 2w + 2=0$$ which gives $$w_{1,2} = -1 \pm i = \sqrt{2} e^{i(\pi \pm \frac{\pi}{4})}$$ Now you got two equations to solve \begin{align} z_1^5 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4})} \\ z_2^5 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4})} \end{align} which gives \begin{align} z_1 &= \sqrt{2} e^{i(\pi + \frac{\pi}{4} + \frac{2k\pi}{5})} \\ z_2 &= \sqrt{2} e^{i(\pi - \frac{\pi}{4}+ \frac{2k\pi}{5})} \end{align} for $k \in \lbrace 0,1,2,3,4 \rbrace$

Answer 2

$$1+z^5=-1\pm i=\sqrt2e^{i(2n\pi+\pi\pm\pi/4)}$$ where $n$ is any integer

$$z=2^{1/10}\text{exp}\left(\dfrac{i\left(2n\pi+\pi\pm\pi/4\right)}5\right)$$ where $0\le n\le4$

Observe that $$\text{exp}\left(\dfrac{i\left(2n\pi+\pi+\pi/4\right)}5\right)\ne\text{exp}\left(\dfrac{i\left(2n\pi+\pi-\pi/4\right)}5\right)$$ as the equality needs $\text{exp}\left(\dfrac{i\pi/2}5\right)=1$

Like How to solve $x^3=-1$?, we can prove that there is no repeated root of the given equation

Now if for $n_1<n_2$ $$\text{exp}\left(\dfrac{i\left(2n_1\pi+\pi+\pi/4\right)}5\right)=\text{exp}\left(\dfrac{i\left(2n_2\pi+\pi+\pi/4\right)}5\right)$$

$\iff\dfrac{2(n_1-n_2)}5$ is a multiple of $2\pi$

$\iff n_1-n_2$ is a multiple of $5$ which is not possible as $0\le n_1<n_2\le4$

Similarly for

$$\text{exp}\left(\dfrac{i\left(2n_1\pi+\pi-\pi/4\right)}5\right)=\text{exp}\left(\dfrac{i\left(2n_2\pi+\pi-\pi/4\right)}5\right)$$