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I have to find the domain of

$\frac{z^2-1}{z^2+z+1}$

So, there can't be any $z$ that makes null the denominator,

$z^2+z+1$ = 0

and after decomposing $z$,

$x^2-y^2+x+2xyi+yi=0$

How to solve that quadratic complex equation?

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you can solve a quadratic equation in a complex variable $z$ the same way you'd solve a quadratic equation in a real variable.. by completing the square or using (appropriately) the quadratic formula. –  Zarrax Feb 15 '11 at 21:23
    
You could compute $(z^2+z+1)(z-1)$ and proceed from there. –  Did Feb 15 '11 at 21:24

2 Answers 2

up vote 5 down vote accepted

Three points:

  1. You did the computation wrong. Writing $z=x+iy$, then $$z^2+z+1 = (x+iy)^2 + (x+iy) + 1 = (x^2-y^2 + x + 1) + i(2xy+y).$$

  2. If $(x^2-y^2+x+1) + i(2xy+y) = 0$, with $x,y$ real numbers, then you need $x^2-y^2+x+1=0$ and $2xy+y=0$. You solve these the way you usually solve equations for real numbers.

    So, for example, you have $0=2xy+y = (2x+1)y$. So either $y=0$ or $2x+1=0$. If $y=0$, then the first equation reduces to $x^2+x+1=0$, which has no real solutions, so there are no solutions with $y=0$. If $y\neq 0$, then $2x+1=0$, so $x=-\frac{1}{2}$. Plugging into the first equation, we get $$0 = \frac{1}{4}-y^2 -\frac{1}{2} + 1 = -y^2 +\frac{3}{4},$$ so you get that $y^2 = \frac{3}{4}$, or $y = \pm\frac{\sqrt{3}}{2}$. So the two solutions are $z=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $z=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.

  3. The acrobatics from step 2 are unnecessary. You don't have to decompose into real and imaginary parts, because the quadratic formula works for complex numbers! (Provided you take complex square roots). Since $$z^2 + z + 1 = \left(z+\frac{1}{2}\right)^2 + \frac{3}{4}$$ (by completing the square), then this is zero if and only if $$z+\frac{1}{2} = \sqrt{-\frac{3}{4}},$$ if and only if $$z = -\frac{1}{2} + \frac{\sqrt{-3}}{2}.$$ You may recognize this as exactly what you get from the quadratic formula applied to $z^2+z+1$, and you may also recognize them as the solutions you get if you go through the contorsions of step 2 above. So just find the two complex square roots of $-3$, and rejoice! (The quadratic formula works even if the coefficients of the quadratic are complex numbers, instead of real numbers).

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Thanks for this very detailed answer! –  Tomas Sironi Feb 15 '11 at 21:37

The maximal domain of definition is,as you imply, the complement in $\mathbb C$ of the set of roots of the polynomial $z^2+z+1$.

Therefore to find the bad points, we need to solve the equation $$z^2+z+1=0.$$ I am pretty sure, now, that you know how to solve quadratic equations, no?

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I know how to. I messed trying to decompose $z$ when is not necessary. Thanks! –  Tomas Sironi Feb 15 '11 at 21:28

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