How do I find the solution to this equation?

Question

$$ z^3=-1 $$

How do I find all of the solutions to this equation? I get so confused when they use $$z$$ instead of writing the complex number like this $$a+bi$$

What do you find confusing about variables that range over complex numbers? If you take a geometrical view then representing $z$ as $a + bi$ here is unhelpful: cubing a complex number cubes its modulus and triples its argument (modulo $2\pi$), so the solutions of $z^3=-1$ have modulus $1$ and argument $\pi/3$, $\pi$ and $5\pi/3$. — Rob Arthan, Feb 07 '20 at 02:00
https://math.stackexchange.com/questions/192742/how-to-solve-x3-1 — lab bhattacharjee, Feb 07 '20 at 04:05

score 3 · Answer 1 · answered Feb 07 '20 at 00:29

3

HINT

\begin{align*} z^{3} = -1 \Longleftrightarrow z^{3} + 1 = 0 \Longleftrightarrow (z+1)(z^{2} - z + 1) = 0 \Longleftrightarrow (z = -1)\vee(z^{2} - z + 1 =0) \end{align*}

answered Feb 07 '20 at 00:29

user0102

21,572

Thank you so much! It was a really good explanation! – Mathomat55 Feb 07 '20 at 20:55
You are welcome. – user0102 Feb 11 '20 at 20:04

score 2 · Answer 2 · answered Feb 07 '20 at 00:37

2

Hint: $$z^3=e^{i \pi} \iff z=e^{i(\pi/3 +2 k \pi/3)}, \quad k=0,1,2.$$

answered Feb 07 '20 at 00:37

Migalobe

238
1
14

Frankly this is more of an answer than a hint :) – 79037662 Feb 07 '20 at 01:07

J. W. Tanner · Answer 3 · 2020-02-07T01:16:09.370

1

Let $z=a+bi$ with $a,b\in\mathbb R$. Assume $z^3=-1$.

$(a+bi)^3=a^3+3a^2bi-3ab^2-b^3i=-1$ means $a^3-3ab^2=-1$ and $3a^2b-b^3=0$.

From the latter equation, $b(3a^2-b^2)=0$, so $b=0$ or $b^2=3a^2$.

If $b=0$, then from $a^3-3ab^2=-1$ we get $a^3=-1$ or $a=-1$.

If $b^2=3a^2$, then from $a^3-3ab^2=-1$ we get $a^3-9a^3=-8a^3=-1$,

so $a^3=\frac18$, so $a=\frac12$, and $b^2=\frac34$.

Can you take it from here?

edited Feb 07 '20 at 01:16

answered Feb 07 '20 at 01:07

J. W. Tanner

60,406

1

Thank you so much!! Yes, I can take it from there! – Mathomat55 Feb 07 '20 at 20:57

score 1 · Accepted Answer · edited Feb 07 '20 at 11:55

Euler's identity is your friend. Euler's identity says that

$e^{i \theta} = \cos(\theta) + i \sin(\theta)$

After some experimentation with different values of $\theta$, one can see that $e^{i\pi} = -1$, and so it becomes clear that we need is for our answer cubed to give us an angle of $\theta = \pi$. Thus, $\theta = \dfrac{\pi}{3}$ and $\theta = \dfrac{5\pi}{3}$ are the other complex angles that you're looking for, as multiplying either of them by $3$ results in an angle that is co-terminal to $\pi$.

score 1 · Answer 5 · answered Feb 07 '20 at 02:05

You can also use a formula for extracting $n$-th roots of complex numbers, which is the following:

$$w_k= \sqrt[n]{\rho} \,\,\cdot \left\lbrace \cos\left( \frac{\theta + 2k\pi}{n} \right) + i\sin\left( \frac{\theta + 2k\pi}{n} \right)\right\rbrace $$

where $k$ is integer, $\rho$ is the distance in respect to the origin in the complex plane and $\theta$ is the argument of the complex number. In your situation, it would be: $\rho$ = 1, $k =0,1,2$, $n = 3$ and $\theta = -\pi$. Then we get

$$w_{0,1}= \frac{1}{2} \mp i\frac{\sqrt3}{2}$$ $$w_2 = -1$$

How do I find the solution to this equation?

5 Answers5