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How to solve $x^3=-1$? I got following: $x^3=-1$

$x=(-1)^{\frac{1}{3}}$

$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...

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  • $\begingroup$ It may be interesting to analyze why this method does not produce all results: $x^3=−1$ so $x.x^2=−1∗1$ so x=−1 and/or $x=+1$ and/or $x=−1$, since x can't be both +1 and -1, so x=−1. $\endgroup$ – NoChance Sep 9 '12 at 8:51
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$x^3+1=0\implies (x+1)(x^2-x+1)=0$

If $x+1=0,x=-1$

else $x^2-x+1=0$ then $x=\frac{1±\sqrt{1-4}}{2\cdot1}=\frac{1±\sqrt3i}{2}$ using the well known Quadratic Formula.

Alternatively, using Euler's identity and Euler's formula like other two solutions, for $x^n=-1=e^{i(2m+1)\pi}$ as $e^{i\pi}=-1$ where $m$ any any integer, $n$ is a natural number .

We know $n$-degree equation has exactly $n$ roots, so the roots of $x^n+1=0$ are $e^{\frac{i(2m+1)\pi}{n}}=\cos\frac{(2m+1)\pi}{n}+i\sin \frac{(2m+1)\pi}{n}$ where $m=0,1,2,...n-1$. It's just customary, not mandatory that we have defined $m$ to assume this range of values, in fact it can assume any $n$ in-congruent values (for example, consecutive values like $r,r+1,...,r+n-1,$ where r is any integer), the reason is explained below.

(1)Using Repeated Root Theorem, let $f(x)=x^n+R$ where $n>1$ and $R ≠ 0\implies x≠0$

So, $\frac{df}{dx}=nx^{n-1}, \frac{df}{dx}=0$ does not have any non-zero root.

Clearly, $f(x)=x^n+R$ can not have any repeated root unless $R=0$

(2)Let $e^{\frac{i(2a+1)\pi}{n}}=e^{\frac{i(2b+1)\pi}{n}}$ $\implies e^{\frac{2i(a-b)\pi}{n}}=1=e^{2k\pi i}$ where $k$ is any integer.

$\implies a-b=kn\implies a≡b\pmod n$, so any $n$ in-congruent values of $m$ will give us essentially the same set of n distinct solutions.

Here $n=3$ so, let's take $m=0,1,2$.

$m=0\implies \cos\frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1+i\sqrt3}{2} $

$m=1\implies \cos \pi+i\sin\pi=-1 $

$m=2\implies \cos\frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac{1-i\sqrt3}{2} $

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  • $\begingroup$ I think the second factor in your first line should be $x^2-x+1$. $\endgroup$ – Daryl Sep 8 '12 at 12:32
  • $\begingroup$ @Daryl, thanks for identifying the lapsus calami. $\endgroup$ – lab bhattacharjee Sep 8 '12 at 12:34
  • $\begingroup$ Hi, where did you got $x^3+1=(x+1)(x^2-x+1)$? I mean also the site, because I found $x^n-1=(x-1)(x^{n-1}+ \dots +1)$ - planetmath.org/encyclopedia/PrimeFactorsOfXn1.html. $\endgroup$ – user2723 Sep 8 '12 at 19:02
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    $\begingroup$ $x^3+y^3=(x)^3+(y)^3=(x+y)^3-3xy(x+y)=(x+y)((x+y)^2-3xy)=(x+y)(x^2+y^2-xy)$ $\endgroup$ – lab bhattacharjee Sep 9 '12 at 3:39
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Let $x=a+bi$, where $a$ and $b$ are real. Then $(a+bi)^3 = -1$. Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$

We can separate the real and imaginary parts of this equation: $$\begin{align} a^3 -3ab^2 & = -1 \\ 3a^2b - b^3 & = 0 \end{align}$$

Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution

$$x =-1.$$

So suppose $b\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \pm a\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$ a^3 - 9a^3 = -1$$ so $a = \frac12.$ Since $b = \pm a\sqrt3$, we have $b=\pm\frac{\sqrt3}2$. This gives us the other two solutions, namely

$$x=\frac12 \pm\!\frac{\sqrt3}2i.$$

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Set $\displaystyle x=re^{i \theta}$. So $\displaystyle r^3e^{i3\theta}= x^3= -1= e^{i \pi}$, hence $r^3=1$ and $3 \theta= \pi [2\pi]$. Finally, $r=1$ and $\displaystyle \theta = \frac{\pi}{3} \left[\frac{2\pi}{3} \right]$ so you get three solutions: $$x_1= e^{i \pi/3}= \frac{1}{2}+ i \frac{\sqrt{3}}{2}, \ x_2=e^{-i 2 \pi /3}= \frac{1}{2} - i \frac{\sqrt{3}}{2}, \ x_3= e^{i \pi}=-1.$$

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  • $\begingroup$ What does $3 \theta = \pi[2 \pi]$ and $\theta = \frac{\pi}{3}[\frac{2 \pi}{3}]$ mean? Do they mean $3 \theta = \pi+2 \pi n$ and $\theta = \frac{\pi}{3}+ \frac{2 \pi}{3} n$? $\endgroup$ – user2723 Sep 13 '12 at 6:47
  • $\begingroup$ It is indeed the correct definition. $\endgroup$ – Seirios Sep 13 '12 at 8:46
  • $\begingroup$ well how do you get $\frac{2 \pi}{3}$, because $\theta = \frac{\pi}{3}, \pi, \frac{5 \pi}{3}, \dots$, but you don't find $\frac{2 \pi}{3}$. $\endgroup$ – user2723 Sep 13 '12 at 11:02
  • $\begingroup$ Indeed, I forgot a minus. Thank you! $\endgroup$ – Seirios Sep 16 '12 at 8:15
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Observe that $(e^{a i})^3 = e^{3 a i}$ and $-1=e^{\pi i}$.

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$$x^3=-1$$

$$x^3+1=0$$

$$(x+1)(x^2+1-x)=0$$

$$x=-1 \quad\text{or}\quad x^2-x+1=0$$

When $$x^2-x+1=0, x= \frac{-1(\pm\sqrt{-3})}{2}$$

$$x= \frac{-1+\sqrt{3}i}{2} \quad\text{and}\quad \frac{-1-\sqrt{3}i}{2}$$

Which is equal to $e^{−i2π/3}$ and $e^{i2π/3}$ respectively.

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The real solution:

$$x^3+1=0<=>$$ $$x^3=-1<=>$$

Take cube roots of both sides:

$$x=-1$$

Complex solution:

$$x^3+1=0<=>$$ $$x^3=-1<=>$$ $$x^3=|-1|e^{arg(-1)i}<=>$$ $$x^3=e^{(\pi +2\pi k)i}<=>$$

(with k is the element of Z)

$$x=\left(e^{(\pi +2\pi k)i}\right)^{\frac{1}{3}}<=>$$ $$x=e^{\left(\frac{1}{3}\pi +\frac{2}{3}\pi k\right)i}$$

(with k goes from 0 to 3 -> k=0-2)

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Complex numbers have a great visual interpretation. $x+ iy$ is visualized as the point in the plane with coordinates $(x,y)$. When you multiply two complex numbers, you "add the angles and multiply the lengths". This visual interpretation of complex number multiplication allows us to find the solutions to $x^3 = -1$ immediately, with no writing. Of course $-1$ is one solution. You probably see another solution already. It's on the unit circle and it makes an angle of 60 degrees with the x-axis. And finally you see the third solution as well.

It's a fact that polynomials of degree $3$ have at most $3$ distinct roots, so we have found all the solutions to $x^3+1=0$.

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