Solve z^8 + 1 = 0 in polar form and factorize the polynomial

Key takeaway: This is a roots-of-unity style problem. The key move is to rewrite $-1$ in polar form and then take eighth roots.

We solve $z^8+1=0\quad\Rightarrow\quad z^8=-1.$

Write $-1$ in polar form: $-1=e^{i(\pi+2\pi m)}\qquad (m\in\mathbb{Z}).$

Taking eighth roots gives $z=e^{i(\pi+2\pi m)/8}=e^{i(2m+1)\pi/8}.$

For the 8 distinct roots, take $m=0,1,2,3,4,5,6,7$: $z_k=e^{i(2k+1)\pi/8},\qquad k=0,1,2,3,4,5,6,7.$

So in polar form the solutions are $\boxed{e^{i\pi/8},\ e^{i3\pi/8},\ e^{i5\pi/8},\ e^{i7\pi/8},\ e^{i9\pi/8},\ e^{i11\pi/8},\ e^{i13\pi/8},\ e^{i15\pi/8}}.$

Hence $z^8+1=\prod_{k=0}^{7}\left(z-e^{i(2k+1)\pi/8}\right).$

A real factorization is obtained by pairing conjugates: $z^8+1=(z^4+\sqrt2 z^2+1)(z^4-\sqrt2 z^2+1).$

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Pitfalls the pros know 👇 A frequent error is writing the roots as $e^{i\pi/8}$ only and forgetting the other seven values. Another trap is using $2\pi k/8$ instead of $(2k+1)\pi/8$; that would solve $z^8=1$, not $z^8=-1$.

What if the problem changes? If the equation were $z^8-1=0$, the roots would be $e^{i2\pi k/8}$ instead. If the question asked for Cartesian form, you would expand $e^{i\theta}=\cos\theta+i\sin\theta$ for each root.

Tags: complex roots, polar form, roots of unity