Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have to find the domain of


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

$z^2+z+1$ = 0

and after decomposing $z$,


How to solve that quadratic complex equation?

share|cite|improve this question
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
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).

share|cite|improve this answer
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?

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.