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This question already has an answer here:

I'm working on this problem and I got stuck, here is my attempt.

Find all $z\in\mathbb{C}$ such that $z^3 + 1 = 0$.

Solution. Since $z$ is a complex number, then we must have $z = x + yi$ where $x,y$ are real numbers, hence $$\left(x + yi\right)^3 = -1$$ or $$x^3 + 3x^2yi - 3xy^2 - y^3i = -1$$ by equality of the real and imaginary part, we obtain the system $$\mbox{$x^3 + 3xy^2 = - 1$ and $3x^2y - y^3 = 0$}$$ and this is where I got stuck, I even made some substitution to eliminate one variable but it didn't help matters.

Please someone tell me what to do next. And please don't try solving it using polar forms, just algebraic manipulation. Thanks.

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marked as duplicate by Dietrich Burde, kingW3, littleO, drhab, Claude Leibovici Jun 28 '17 at 9:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See math.stackexchange.com/questions/192742/how-to-solve-x3-1 $\endgroup$ – lab bhattacharjee Jun 28 '17 at 8:50
  • $\begingroup$ What do you mean by "polar forms" and why can't we do it that way? If you know the geometric interpretation of complex number multiplication, you can immediately find the solutions visually. $\endgroup$ – littleO Jun 28 '17 at 8:52
  • $\begingroup$ He meant polar co-ordinates, and this was perhaps because he marked the question as pre-calculus $\endgroup$ – user456218 Jun 28 '17 at 8:54
  • $\begingroup$ @Burde, it is not a duplicate post as in this post $z$ is a complex number. $\endgroup$ – Icosahedron Jun 28 '17 at 8:54
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    $\begingroup$ @Icosahedron In the linked post $x$ is also complex. $\endgroup$ – kingW3 Jun 28 '17 at 8:59
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Hint. Note that $z=-1$ is a solution of $z^3+1=0$. It follows that $z^3+1=(z+1)(z^2-z+1)$. Hence, it remains to solve in $\mathbb{C}$ the quadratic equation $$z^2-z+1=\left(z-\frac{1}{2}\right)^2+\frac{3}{4}=0.$$ Can you take it from here?

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  • $\begingroup$ $z = -1$ ? But z is a complex number, how would that bail me out? $\endgroup$ – Icosahedron Jun 28 '17 at 8:58
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    $\begingroup$ @Icosahedron $z=-1=-1+0\cdot i$ IS a complex number! See my edited answer. $\endgroup$ – Robert Z Jun 28 '17 at 9:05
  • $\begingroup$ Thanks a bunch. I've solved it completely and got the correct answer. $\endgroup$ – Icosahedron Jun 28 '17 at 9:11

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