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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?

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    $\begingroup$ 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
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    $\begingroup$ Hint: write it as $\,z^3-i^3=0\,$. $\endgroup$ – dxiv Sep 28 '18 at 5:02
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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.

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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.

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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 $k=0,1,2$

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