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Solve the equation $z^2 + z + 1 = 0$ for $z = (x,y)$ by writing $(x,y) (x,y) + (x,y) + (1,0) = (0,0)$ and then solving a pair of simultaneous equations in $x$ and $y$.

My main difficulty is finding the answer is using the suggestion in that book that there isn't a real solution implies that $y \neq 0$, but I will work out the rest of this solution just to be sure that I grasp the logic.

First, using the fact that we have some complex variable $z = (x,y)$ and that we can represent a purely real number as $x = (x,0)$, meaning that $1 = (1,0)$, we can reason, as the problem suggests, that \begin{equation} z^2 + z + 1 = 0 \implies (x,y)(x,y) + (x,y) + (1,0) = (0,0) \end{equation} Using multiplication and addition of complex variables, we get \begin{equation} \left(x^2 - y^2, 2xy \right) + \left(x + 1, y\right) = (0,0) \implies \left(x^2 - y^2 + x + 1, 2xy + y\right) = (0,0) \end{equation} which then implies that $x^2 - y^2 + x + 1 = 0$ and $2xy + y = 0$. From this second equation, we get $2xy + y = y(2x + 1) = 0$, which implies $y = 0$ or $2x + 1 = 0 \implies x = - \frac{1}{2}$. Plugging $y = 0$ into the first equation gives us $x^2 - 0^2 + x + 1 = x^2 + x + 1 = 0$, which doesn't have any real solutions.

Now, the suggestion states that the lack of a real solution solution, $x$, when $y = 0$ is justification for throwing out this value of $y$. If I'm not mistaken, and this may well be completely obvious, the logic for this is that we're trying to solve for some value $z = (x,y)$, so we obviously require a real solution even if our result is purely imaginary, though no such $x$ exists. I still I think am not completely convinced of this, as I'm not totally sure why we couldn't reason that we have a purely imaginary solution, i.e., $x = 0$.

But, I think my confusion is in the above step. After this, the algebra follows and gives me the same answer as the textbook, $z = \left(-\frac{1}{2}, \pm \frac{\sqrt{3}}{2}\right)$.

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  • $\begingroup$ If $x=0$ you have to find $y$ by solving the simultaneous equations $-y^2+1=0$ and $y=0$. Try it! $\endgroup$ – David Jul 3 '18 at 1:00
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Note that, for a complex $z = (x,y)$, $x$ and $y$ are both reals. The suggestion is not to require $z$ to be real. We need to find $z$, which is a complex number, and, being a complex number, it has both its real and imaginary components real numbers. Lack of a real solution of $x^2+x+1=0$ suggests that $y \neq 0$.

An alternative way to solve $z^2 + z +1=0$ is to use the famous completing square method. That is, $$z^2+z+1=\left(z+\frac{1}{2}\right)^2+\frac{3}{4}=0 \implies z+\frac{1}{2}=\pm \frac{\sqrt{3}}{2} i.$$

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You can not have a pure imaginary solution for $$z^2 + z + 1 = 0$$ because in that case $z=iy$ where y is a real number.

That implies $$ -y^2+iy+1=0$$ $$iy=y^2-1$$ which is not possible.

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