Solve in $C$ : $P(z)=z^4+2z^3+5z^2+4z+1=0$ where $P(i-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$ Question solve in $C$ : 
$P(z)=z^4+2z^3+5z^2+4z+1=0$ 
where $P(i+-\frac{1}{2}+i\frac{\sqrt 3}{2})=0$ 
My attempt : 
Let $\lambda=i-\frac{1}{2}+i\frac{\sqrt 3}{2}$
$(2\lambda+1)^2=(i(2+\sqrt 3))^2$
$(2\lambda+1)^2+7)^2=(-4\sqrt 3)^2$
$({\lambda}^2+\lambda+2)^2=3$
So we find : 
$\lambda^2++2\lambda^3+5\lambda^2+4\lambda+1=0$
But which step !?  can be find  all root of P(z) !!

If $$z=-\frac{1}{2}-i-\frac{1}{2}\sqrt{3}i$$ is one solution then is also $$z=-\frac{1}{2}+i+\frac{1}{2}\sqrt{3}i$$ is a solution.
