Solving $z^4 + 2z^3 + 6z - 9 = 0$ I'm trying to solve $z^4 + 2z^3 + 6z - 9 = 0$.
$z$ is a complex number.
I usually can solve those equations when they are of second degree.
I don't know what to do, breaking out $z$ doesn't help...
EDIT: Sorry I forgot to mention that $z$ has a solution where $\Re(z) = 0$.

The Quartic Formula can be very useful to solve this.
But there is an alternate way, which my lecturer suggested me : Summing and Subtracting some powers of $z$ to make expression factorable.
For example $z^4+2z^3+6z-9=0$ can be factorized as:

$z^4+(3z^3-z^3)+(3z^2-3z^2)+(9z-3z)-9=0$
  $(z^4+3z^3+3z^2+9z)-(z^3+3z^2+3z+9)=0$
  $z(z^3+3z^2+3z+9)-(z^3+3z^2+3z+9)=0$
  $z(z^2(z+3)+3(z+3))-(z^2(z+3)+3(z+3))=0$
  $(z-1)(z+3)(z^2+3)=0$.

Thus, $z^2+3=0$ or $z=\pm i \sqrt{3}$ are the required solutions for your given conditions.
