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) !!

You are given a polynomial $P(x)$ in integer coefficients and an expression
of one of the roots $$\lambda = i - \frac12 + i \frac{\sqrt{3}}{2} = \sqrt{-1} - \frac12 + \frac12 \sqrt{-1}\sqrt{3}$$
which consists of a bunch of square roots of rational numbers. 
When you substitute $\lambda$ into $P(x)$, you get $P(\lambda) = 0$. If you follow the expansion of $P(\lambda)$ step by step, you will notice all the square roots in $\lambda$ get squared out, become rational number and disappear in final output. The final expression doesn't depend on the signs of square roots. It only depends on squaring a square root give you back the original number.
This has a very useful implication. If you flip the signs of any of the square roots
in $\lambda$, you get another root!
Since there are two independent square roots, $\sqrt{-1}$ and $\sqrt{3}$ in $\lambda$. Flipping their signs independently give you totally $4$ roots.
$$
\begin{align}
\lambda_{++} &= +i - \frac12 + i\frac{\sqrt{3}}{2}\\
\lambda_{+-} &= +i - \frac12 - i\frac{\sqrt{3}}{2}\\
\lambda_{-+} &= -i - \frac12 - i\frac{\sqrt{3}}{2}\\
\lambda_{--} &= -i - \frac12 + i\frac{\sqrt{3}}{2}
\end{align}$$
Since $P(x)$ is a polynomial of degree $4$, these are all the roots of $P(x)$.
