Complex number equation solving Can somebody help me to solve this equation?
$$(\frac{iz}{2+i})^3=-8$$ ?
I'm translating this into
$(\frac{iz}{2+i})=-2$
But i recon it's wrong ...

You are in the right way 
$$\frac { iz }{ 2+i } =\sqrt [ 3 ]{ 8\cdot \left( -1 \right)  } =-2\left( \cos { \frac { k\pi  }{ 3 } +i\sin { \frac { k\pi  }{ 3 }  }  }  \right) ,k=0,1,2$$  check for instance  $k=0$ 
$$\frac { iz }{ 2+i } =-2\\ iz=-4-2i\\ z=\frac { -4-2i }{ i } =\frac { -4i+2 }{ { i }^{ 2 } } =-2+4i$$
