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$$ z^3=i $$

The problem simply states to solve for $z$, but I know that there is some concept to be practiced here about the nth roots of unity. I'm just beginning to learn this concept so I didn't really know how to go about solving this, but below is my attempt:

$Let \ z=x+iy$ $$ z=i^{\frac{1}{3}} = -i $$

$$ r=\sqrt{0^2+(-1)^2}=1 $$

$$ \theta=-\frac{\pi}{2} $$

$$ z=(1)e^{i(\frac{-\pi}{2})}=e^{\frac{-\pi}{2}i} $$

whatwhatwhat
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    If you care, note that for all complex numbers $z$ it holds that $z^3=1\iff (z-1)(z^2+z+1)=0$. – Git Gud Jan 23 '16 at 01:25
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    Write $1$ in polar form. i.e. $1=e^{2k\pi}$ –  Jan 23 '16 at 01:26
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    How did you go from $z=x+iy$ to $z=i^{\frac13}$? – Cameron Buie Jan 23 '16 at 01:55
  • @GitGud from that I would extrapolate that the roots of $z$ are $z=1, -1^{\frac{1}{3}}, and -1^{\frac{2}{3}}$. But what does this mean in regards to the original question? – whatwhatwhat Jan 23 '16 at 01:57
  • @CameronBuie my mistake, the original problem is supposed to be $z^3=i$, not $z^3=1$. But I arrived at that because if $z$ is being cubed, then by taking the cube root of both sides I can get $z$ by itself. – whatwhatwhat Jan 23 '16 at 01:58
  • I posted an answer to the original problem using an elementary technique, just replace $1\mapsto i$ and do the same: equate real and imaginary parts. – YoTengoUnLCD Jan 23 '16 at 02:01
  • Related : http://math.stackexchange.com/questions/1621969/de-moivres-theorem-question-and-complex-numbers – lab bhattacharjee Jan 23 '16 at 03:18

4 Answers4

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This answer was meant for the original problem, the one that had $z^3=1$, the new one can be solved in the exact same manner, replacing $1\mapsto i$.

This method is ugly but requires the fewest theoretical knowledge.

Let $z=x+iy. x,y\in\Bbb R$, then

$$(x+iy)^3=1\iff (x^3-3xy^2)+(3x^2y-y^3)i=1$$

Then $$ x^3-3xy^2=1\\ 3x^2y-y^3=0\iff 3x^2=y^2 $$

Replace in first

$$ x^3-3x(3x^2)=1\iff -8x^3=1\iff x=-\frac 1 2 $$

Then $$y^2=\frac 3 4\rightarrow y=\pm\frac {\sqrt{3}} 2$$

YoTengoUnLCD
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  • But what is this supposed to tell me about $z$? – whatwhatwhat Jan 23 '16 at 17:31
  • Well... You know that $z=x+iy$. If you know $x,y$ don't you know $z$? – YoTengoUnLCD Jan 23 '16 at 17:34
  • Ah ok, so $z=x+iy$ was a valid assumption then? The problem only states what I've written in the first line and the following paragraph. Everything after that was only what I assumed might be the correct way to go about solving this. But if you also agree to let $z=x+iy$ in order to solve this problem then that works for me. – whatwhatwhat Jan 23 '16 at 17:59
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    It's not an assumption at all: If you're given the problem "Find all $z\in \Bbb C$ such that...", you have, from the definition of the complex numbers, that there must exist some $x,y$ real, such that $z=x+iy$. – YoTengoUnLCD Jan 23 '16 at 19:09
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Hint: Use De Moivre's formula.

$z^3 = 1cis(\pi /2)$

$z_k = cis(\pi / 6 +2\pi k / 3)$ Solve for $k=0,1,2$

Ilan Aizelman WS
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They are said to be cube roots of unity so $z^3-1=0$ so $(z-1)(z^2+z+1)=0$ solving $z=1,\omega ,\omega ^2$ where $\omega =\frac{-1+{\sqrt{3}i}}{2}$ and $\omega^2 = \frac{-1-{\sqrt{3}i}}{2}$

Archis Welankar
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$$z^3=i\Longleftrightarrow$$ $$z^3=|i|e^{\arg(i)i}\Longleftrightarrow$$ $$z^3=e^{\frac{\pi i}{2}}\Longleftrightarrow$$ $$z=\left(e^{\left(2\pi k+\frac{\pi}{2}\right)i}\right)^{\frac{1}{3}}\Longleftrightarrow$$ $$z=e^{\frac{1}{3}\left(2\pi k+\frac{\pi}{2}\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-2$

So the solutions are:

$$z_0=e^{\frac{1}{3}\left(2\pi\cdot0+\frac{\pi}{2}\right)i}=e^{\frac{\pi i}{6}}$$ $$z_1=e^{\frac{1}{3}\left(2\pi\cdot1+\frac{\pi}{2}\right)i}=e^{\frac{5\pi i}{6}}$$ $$z_2=e^{\frac{1}{3}\left(2\pi\cdot2+\frac{\pi}{2}\right)i}=-i$$

Jan Eerland
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