I'm trying to find all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$.

I've figured out that

$$(a + ib)^3 = -i \Longleftrightarrow a + ib = \sqrt[3]{-i} \Longleftrightarrow a + ib = -i$$

and also that

$$(a+ib)^3 = a^3 + 3a^2bi + 3ab^2i^2 +i^3b^3$$

but I'm not sure how to proceed from here.

I think I'm supposed to use synthetic division to find the other roots, but not sure what factor to use. I tried $a + ib + i$ and that didn't work.

Any ideas?

  • 1
    $\begingroup$ Find $z$ in the form $re^{i\theta}$ first, then convert to $a+ib$ form. $\endgroup$ – David Sep 28 '18 at 4:46
  • $\begingroup$ Let $w=\frac{z}{i}$, then the equation becomes $w^3=1$. If you are familiar with cube roots of unity, then $w=1,\omega,\omega^2$, so $z=i,i\omega,i\omega^2$. $\endgroup$ – Anurag A Sep 28 '18 at 5:00
  • 1
    $\begingroup$ Hint: write it as $\,z^3-i^3=0\,$. $\endgroup$ – dxiv Sep 28 '18 at 5:02

Without using polar representation:

Note that

$i^3 = i^2 i = (-1)i = -i; \tag 1$

thus $i$ is a solution of

$z^3 = -i, \tag 2$

which may be re-written as

$z^3 + i = 0; \tag 3$

we may seek the other zeroes of (2)-(3) via synthetic division; we seek

$z - i \overline{)z^3 + i}; \tag 4$

we have:

$z \; \text{into} \; z^3 \; \text{yields} \; z^2; \tag 5$

$z^2 \times (z - i) = z^3 - iz^2; \tag 6$

$z^3 + i - (z^3 - iz^2) = iz^2 + i = i(z^2 + 1) = i(z + i)(z - i) ; \tag 7$

now we can skip to the chase by noticing that

$(z - i) \overline{)i(z + i)(z - i)} = i(z + i) = iz - 1; \tag 8$

therefore the quotient should be

$(z - i) \overline{)z^3 + i} = z^2 + iz - 1; \tag 9$

we check:

$(z - i)(z^2 + iz - 1) = z^3 + iz^2 - z - iz^2 + z + i = z^3 + i; \tag{10}$

the quotient (4) is thus the quadratic $z^2 + iz - 1$; we use the quadratic formula:

$z = \dfrac{1}{2} (-i \pm \sqrt{i^2 - 4(1)(-1)}) = \dfrac{1}{2}(-i \pm \sqrt 3), \tag{11}$

which are easily checked to satisfy (3); I leave that simple task to my readers. As well as the tasks of converting our roots to totally proper $a + bi$ form.


Let $z=a+ib$. You have \begin{align*} (a+ib)^3 & = -i\\ a^3 + 3a^2bi - 3ab^2 -ib^3 & =-i\\ (a^3-3ab^2)+i(3a^2b-b^3) & =-i \end{align*} Equate the real and imaginary parts on both sides to get \begin{align*} a^3-3ab^2 =a(a^2-3b^2)& =0\\ 3a^2b-b^3 =b(3a^2-b^2)& =-1 \end{align*} From the first of these equations, we get either $a=0$ or $a^2=3b^2$.

With $a=0$, from the second equation, we get $b^3=1$. Since $b \in \mathbb{R}$, so $b=1$. This gives $\color{red}{z=i}$ as a solution.

With $a^2=3b^2$, from the second equation, we get $8b^3=-1$. Since $b \in \mathbb{R}$, so $b=-\frac{1}{2}$. Consequently we get $a=\pm\frac{\sqrt{3}}{2}$ This gives $\color{red}{z=\frac{\sqrt{3}}{2}+\frac{i}{2}}$ and $\color{red}{z=\frac{-\sqrt{3}}{2}+\frac{i}{2}}$ as other two solutions.


To find $z$ such that $z=(-i)^\frac{1}{3}$. So write $-i=\cos(\frac{\pi}{2})+i \sin (-\frac{\pi}{2})$

$z^3=-i$ implies

$z=(-i)^\frac{1}{3}=\Big(\cos(\frac{\pi}{2})+i \sin (-\frac{\pi}{2})\Big)^\frac{1}{3}=\Big(\cos(2k\pi+\frac{\pi}{2})+i \sin (2k\pi-\frac{\pi}{2})\Big)^\frac{1}{3}$

which is same as $$\cos\Big(2k+\frac{1}{2}\Big)\frac{\pi}{3}+i \sin \Big(2k-\frac{1}{2}\Big)\frac{\pi}{3}$$ where $k=0,1,2$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.