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Solve the equation $z^7=-1$

My attempt:

$$z=x+yi$$

$$(x+yi)^7+1=0$$

$$(x^2+2yi-y^2)^3(x+yi)+1=0$$

but now it's start to look ugly.

I'm sure that there is a simple way

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    $\begingroup$ Do you know the polar form of complex numbers? $\endgroup$
    – Daniel Fischer
    Commented Nov 3, 2015 at 14:37
  • $\begingroup$ @DanielFischer Yes, $z=re^{i\theta}$ $\endgroup$
    – user238874
    Commented Nov 3, 2015 at 14:38
  • $\begingroup$ @DanielFischer Sorry, it is r$(\cos(\theta)+i\sin(\theta))$ $\endgroup$
    – user238874
    Commented Nov 3, 2015 at 14:40
  • $\begingroup$ Both work, but $re^{i\theta}$ is more convenient (here at least). If you write $z$ in that form, what do you get from $z^7 = -1$? $\endgroup$
    – Daniel Fischer
    Commented Nov 3, 2015 at 14:42
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    $\begingroup$ It's $\tan\theta=\frac{y}{x}$, if $x\neq 0$. But that doesn't quite determine $\theta$, since $\tan$ is $\pi$-periodic. Anyway, if $r^7 e^{7i\theta}=-1$, then $r^7=\lvert r^7 e^{7i\theta}\rvert =\lvert-1\rvert=1$, which implies $r=1$. Next, we know $e^{i\varphi}= -1\iff \varphi = (2k+1)\pi$ for some $k\in \mathbb{Z}$, so we must have $7\theta = (2k+1)\pi$, and $\theta = \frac{2k+1}{7}\pi$ for some $k\in \mathbb{Z}$. Since $e^{2\pi i} = 1$, two values of $k$ give the same $z$ when they differ by a multiple of $7$, so you have the seven values $z_k=e^{(2k+1)/7\pi}$ for $0\leqslant k\leqslant 6$. $\endgroup$
    – Daniel Fischer
    Commented Nov 3, 2015 at 16:09

3 Answers 3

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We can say that $$z^7=-1$$ or,$$z^7=\cos\pi+i\sin\pi$$ or,$$z^7=\cos(4n+2)\pi+i\sin(4n+2)\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$ or,$$z=\left[\cos(4n+2)\pi+i\sin(4n+2)\pi\right]^{\frac{1}{7}} \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$ or,$$z=\cos(\frac{4n+2}{7})\pi+i\sin(\frac{4n+2}{7})\pi \,\ \text{where} \,\ n=0,1,2,3,4,5,6$$

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$$z^7=-1\Longleftrightarrow$$ $$z^7=|-1|e^{\arg(-1)i}\Longleftrightarrow$$ $$z^7=1\cdot e^{\pi i}\Longleftrightarrow$$ $$z^7=e^{\pi i}\Longleftrightarrow$$ $$z=\left(e^{\left(\pi+2\pi k\right)i}\right)^{\frac{1}{7}}\Longleftrightarrow$$ $$z=e^{\frac{1}{7}\left(\pi+2\pi k\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-6$

So the solutions are:

$$z_0=e^{\frac{1}{7}\left(\pi+2\pi\cdot 0\right)i}=e^{\frac{\pi}{7}i}$$ $$z_1=e^{\frac{1}{7}\left(\pi+2\pi\cdot 1\right)i}=e^{\frac{3\pi}{7}i}$$ $$z_2=e^{\frac{1}{7}\left(\pi+2\pi\cdot 2\right)i}=e^{\frac{5\pi}{7}i}$$ $$z_3=e^{\frac{1}{7}\left(\pi+2\pi\cdot 3\right)i}=e^{\pi i}=-1$$ $$z_4=e^{\frac{1}{7}\left(\pi+2\pi\cdot 4\right)i}=e^{-\frac{5\pi}{7}i}$$ $$z_5=e^{\frac{1}{7}\left(\pi+2\pi\cdot 5\right)i}=e^{-\frac{3\pi}{7}i}$$ $$z_6=e^{\frac{1}{7}\left(\pi+2\pi\cdot 6\right)i}=e^{-\frac{\pi}{7}i}$$

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HINT: Try using this,(one of the solution)
$$ z^7=e^{i\pi} \\z=e^{i\frac{\pi}{7}} $$

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  • $\begingroup$ This isn't the only solution. $\endgroup$
    – Wojowu
    Commented Nov 3, 2015 at 15:01
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    $\begingroup$ @Wojowu I didn't say this is the only solution. And certainly this answer does not deserve a down vote. $\endgroup$
    – Sathyaram
    Commented Nov 3, 2015 at 15:04
  • $\begingroup$ It wasn't me who downvoted. I meant that the way you formulated this hint suggests that you are claiming this is the only solution. $\endgroup$
    – Wojowu
    Commented Nov 3, 2015 at 15:06

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