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Solve:

$$z^6 + 1 = 0$$

That lie in the top region of the plane.

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We know that:

$$(z^2 + 1)(z^4 – z^2 + 1) = 0$$

$$z = -i, i$$

We need to solve:

$$((z^2)^2 – (z)^2 + 1) = 0$$

$$z = \frac{1 \pm \sqrt{-3}}{2}$$

But this is incorrect. How to do this then?

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$$z^6=-1=e^{(2n+1)\pi i}$$ where $n$ is any integer

$$\implies z=e^{\dfrac{(2n+1)\pi i}6}=\cos\dfrac{(2n+1)\pi}6+i\sin\dfrac{(2n+1)\pi}6$$ where $0\le n\le 5$

Top region of the plane, $\implies$ the ordinate has to be $>0$

$\implies\sin\dfrac{(2n+1)\pi}6>0\implies0<\dfrac{(2n+1)\pi}6<\pi\iff0<2n+1<6\implies-.5< n<2.5$

$\implies n=0,1,2$

Only your last line is incorrect. What you should write is

$$z^2 = \frac{1\pm\sqrt{-3}}{2}$$

The simplest way of solving this equation is the method based on DeMoivre’s Formula that Lab Bhattacharjee outlined.

That said, you can make your method work. You found the roots $z \pm i$ by setting the factor $z^2 + 1$ equal to zero. As Rasolnikov and 5xum noted, you should have obtained

$$z^2 = \frac{1 \pm \sqrt{-3}}{2}$$

when you set the factor $z^4 – z^2 + 1$ equal to zero.

Let $z = a + bi$, with $a, b \in \mathbb{R}$. Then

\begin{align*}

z^2 & = \frac{1 \pm i\sqrt{3}}{2}\\

(a + bi)^2 & = \frac{1 \pm i\sqrt{3}}{2}\\

a^2 + 2abi – b^2 & = \frac{1 \pm i\sqrt{3}}{2}

\end{align*}

Equating real and imaginary parts yields

\begin{align*}

a^2 – b^2 & = \frac{1}{2}\tag{1}\\

2ab & = \pm\frac{\sqrt{3}}{2}\tag{2}

\end{align*}

Solving equation 2 for $b$ yields

$$b = \frac{\pm\sqrt{3}}{4a}\tag{3}$$

Substituting this expression in equation 1 yields

\begin{align*}

a^2 – \frac{3}{16a^2} & = \frac{1}{2}\\

16a^4 – 3 & = 8a^2\\

16a^4 – 8a^2 & = 3\\

16a^4 – 8a^2 + 1 & = 4 && \text{complete the square}\\

(4a^2 – 1)^2 & = 4\\

4a^2 – 1 & = \pm 2\\

4a^2 & = 3 && \text{since $a \in \mathbb{R}$}\\

a^2 & = \frac{3}{4}\\

a & = \pm \frac{\sqrt{3}}{2}

\end{align*}

Substituting this expression into equation 3 yields the four roots

$$z = \pm\frac{\sqrt{3}}{2} \pm \frac{1}{2}i$$

of the equation $z^4 – z^2 + 1 = 0$. As you can check, these roots correspond to the values $n = 0, 2, 3, 5$ in the formula Lab provided.

In order to solve $\sqrt[2n]{-1}$:

- Draw the unit circle
- Draw the first solution, which is obviously $0+1i=\cos(\frac{\pi}{2})+\sin(\frac{\pi}{2})i$
- Repeat $2n-1$ times: find the next solution by rotating the previous solution $\frac{\pi}{n}$ radians

For example, $\sqrt[6]{-1}$:

- $\cos(\frac{ 3\pi}{6})+\sin(\frac{ 3\pi}{6})i$
- $\cos(\frac{ 5\pi}{6})+\sin(\frac{ 5\pi}{6})i$
- $\cos(\frac{ 7\pi}{6})+\sin(\frac{ 7\pi}{6})i$
- $\cos(\frac{ 9\pi}{6})+\sin(\frac{ 9\pi}{6})i$
- $\cos(\frac{11\pi}{6})+\sin(\frac{11\pi}{6})i$
- $\cos(\frac{13\pi}{6})+\sin(\frac{13\pi}{6})i$

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