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I'm asked to evaluate $$\sqrt[3]{-i}$$ I suppose $\sqrt[3]{-i}=(a+bi)$ $$\implies (a+bi)^3=-i$$ $$\implies \Im \left( (a+bi)^3 \right) =\Im \left( (-i) \right)$$ $$\implies 3a^2b-b^3=-1$$ Now how am I supposed to find $a$,$b$? Aren't there infinitely of them instead of just three?

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  • $\begingroup$ There cannot be infinitely many of them, because $\sqrt[3]{-i}$ are the roots of the complex polynomial $x^3 = -i$, and the fundamental theorem of algebra, says that this will have exactly three complex roots. Do you know what the polar form / polar coordinates are of a complex number? $\endgroup$ – астон вілла олоф мэллбэрг Sep 3 '18 at 3:45
  • $\begingroup$ @астонвіллаолофмэллбэрг For every $a$ there is a corresponding $b$. $\endgroup$ – user588826 Sep 3 '18 at 3:49
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    $\begingroup$ But what about $\mathcal{R}( (a+bi)^3 ) = \mathcal{R}(-i)$ ? It gives you another equation, so there is not infinitely many. $\endgroup$ – Ahmad Bazzi Sep 3 '18 at 3:53
  • $\begingroup$ You have $3a^2b - b^3 = 1$, but then you also have $a^3 - 3ab^2 = 0$, by equating the real parts. One of these equations may have infinitely many solutions, but now we have two simultaneous equations. $\endgroup$ – астон вілла олоф мэллбэрг Sep 3 '18 at 3:57
  • $\begingroup$ @астонвіллаолофмэллбэрг Can we deduce something from both of them? $\endgroup$ – user588826 Sep 3 '18 at 3:59
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Let us start as you did:
$\sqrt [3] {-i}= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$

Therefore we get $2$ equations:
$a^3-3ab^2=0 \ \ \ \ldots(1)$
$3a^2b-b^3=-1 \ldots (2)$

Solving for $(1)$:
$a(a^2-3b^2)=0$
$\therefore a(a-b\sqrt3)(a+b\sqrt 3)=0$

So:
$a=0$
$a=b\sqrt3$
$a=-b\sqrt 3$

Solving for $(2)$:
$3a^2b-b^3=-1$

When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$

When $a=\sqrt {3b^2}$:
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=\frac{\sqrt 3}2$

When $a=-\sqrt {3b^2}$
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=-\frac{\sqrt 3}2$

So your $3$ solutions are:
$(a,b)=(0,1);(\frac{\sqrt 3}2,-\frac 12);(-\frac{\sqrt 3}2,-\frac 12)$

I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!

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You can solve this as \begin{equation} (-i)^{\frac{1}{3}} = ( e^{i \frac{3\pi}{2} + i2k\pi})^{\frac{1}{3}} \end{equation} which gives us three distinct roots $e^{i z_k}$, for $k = 0,1,2$, where \begin{align} z_0 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(0)\pi}{3} = \frac{\pi}{2} \\ z_1 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(1)\pi}{3} = \frac{7\pi}{6} \\ z_2 &= \frac{1}{3}\frac{3\pi}{2} + \frac{2(2)\pi}{3} = \frac{11\pi}{6} \\ \end{align} If you insist on solving it your way, then \begin{equation} (a+bi)^3 = -i \end{equation} means \begin{equation} a^3 + 3a^2bi - 3ab^2 - b^3i = -i \end{equation} which means \begin{align} a^3 - 3ab^2 &= 0 \\ 3a^2b - b^3 &= -1 \end{align} which is \begin{align} a^2 - 3b^2 &= 0 \\ 3a^2b - b^3 &= -1 \end{align} or \begin{align} (a - \sqrt{3} b)(a + \sqrt{3} b) &= 0 \\ 3a^2b - b^3 &= -1 \end{align} The first equation suggests either \begin{equation} a = \pm \sqrt{3} b \end{equation} Replacing this in the second equation gives \begin{equation} 3(\sqrt{3} b)^2b - b^3 = -1 \end{equation} which is \begin{equation} 9b^3 - b^3 = -1 \end{equation} i.e. \begin{equation} b = -\frac{1}{2} \end{equation} This will give us \begin{equation} a = \pm \sqrt{3} (-\frac{1}{2}) \end{equation} which means we get two solutions \begin{align} (a_1,b_1) &= (-\sqrt{3},-\frac{1}{2}) \\ (a_2,b_2) &= (\sqrt{3},-\frac{1}{2}) \\ \end{align} which are actually what we found before, i.e. \begin{align} a_1 + ib_1 &= e^{i z_1} \\ a_2 + ib_2 &= e^{i z_2} \\ \end{align} Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that \begin{align} a_0 + ib_0 &= e^{i z_0} \end{align}

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When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:

$$z^3=-i$$

Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $\frac{3\pi}{2}+2k\pi, k\in \text{Z}$ (this is pointing downwards). The possible angles are:

$$\frac{\pi}{2}+\frac{2k\pi}{3}$$

So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2\pi$). The solutions are:

$$z=\cos{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)} + i\sin{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)}, k\in\text{{0,1,2}}$$

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  • $\begingroup$ $\LaTeX \text{ Tip}:$ Use \sin x and \cos x to obtain $\sin x \text{ and } \cos x.$ $\endgroup$ – Mohammad Zuhair Khan Oct 21 '18 at 6:58
  • $\begingroup$ I edited with better Latex $\endgroup$ – Villa Oct 22 '18 at 4:21
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Alt. hint:   the cube roots of $\,-i\,$ are the solutions to $\,z^3=-i \iff z^3+i=0\,$. Using that $\,i = -i^3\,$ and the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, the latter equation can be written as:

$$ 0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)\big(-(iz)^2+iz-1\big) $$

The first factor gives the (obvious) solution $\,z=i\,$, and the second factor is a real quadratic in $\,iz\,$ with roots $\,iz = (1 \pm i \sqrt{3})/2 \iff z = \ldots$

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It is easier to find the cube roots of -i using the polar form.

Note that $-i = e^{3i\pi /2}$ thus the cube roots are $e^{i\pi/2},e^{i\pi/2+2i\pi/3}, e^{i\pi/2+4i\pi/3}$

You can easily find these roots in $a+bi $ form if you wish to do so.

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