# Finding all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$

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?

• Find $z$ in the form $re^{i\theta}$ first, then convert to $a+ib$ form. – David Sep 28 '18 at 4:46
• 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 $z=i,i\omega,i\omega^2$. – Anurag A Sep 28 '18 at 5:00
• Hint: write it as $\,z^3-i^3=0\,$. – 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:

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

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

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

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

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 $$k=0,1,2$$