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) !!
 A: 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 $P(x)$.
A: 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.
A: Since $P$ has real coefficients, then $z=-\frac12-i-\frac{\sqrt{3}}2i$ is also a solution to $P(z)=0.$ This means that both $z+\frac12-i-\frac{\sqrt{3}}2i$ and $z+\frac12+i+\frac{\sqrt{3}}2i$ are factors of $P(z),$ and so \begin{eqnarray}\left(z+\frac12-i-\frac{\sqrt{3}}2i\right)\left(z+\frac12+i+\frac{\sqrt{3}}2i\right) &=& z^2+z+\left(\frac12-i-\frac{\sqrt{3}}2i\right)\left(\frac12+i+\frac{\sqrt{3}}2i\right)\\ &=& z^2+z+\left(\frac12\right)^2+\left(1+\frac{\sqrt3}2\right)^2\\ &=& z^2+z+\frac14+1+\sqrt3+\frac34\\ &=& z^2+z+2+\sqrt3\end{eqnarray} is also a factor of $P(z).$ Thus, $$z^4+2z^3+5z^2+4z+1=(z^2+z+2+\sqrt3)(az^2+bz+c)$$ for some $a,b,c.$ Expand the right-hand side, and see if you can find $a,b,c.$ From there, you can use the quadratic formula on $az^2+bz+c=0$ to fully solve $P(z)=0.$
A: There are two ways I spotted for doing this quite quickly. Since the equation has real coefficients, the complex conjugate of any root is also a root, so that gives you two roots (the one you have and its conjugate). The remaining two roots will be roots of a quadratic with real coefficients, and you can use the standard (vieta) formulae for the sum and product of the roots of the quartic to work out the quadratic for the other roots via the sum and product of the unknown roots.
The second is more accidental and less systemic, but easier if you spot it. You can rewrite the equation you have as $$0=z^4+2z^3+z^2+4z^2+4z+1=(z^2+z)^2+(2z+1)^2=(z^2+z+2iz+i)(z^2+z-2iz-i)$$which is a simple use of the difference of two squares formula in the complex number context. This gives two quadratics with complex coefficients to solve, but one is the conjugate of the other, so the two quadratics have conjugate roots. All you have to do is to identify which of these has the root you have already been given (a simple computation - if the first one doesn't the second must - though worth doing the second to confirm arithmetic). Then you can use the sum of the roots formula for the quadratic to identify the second root.

Actually from your attempt you have $({\lambda}^2+\lambda+2)^2=3$, which gives you $(\lambda^2+\lambda+2+\sqrt 3)(\lambda^2+\lambda+2-\sqrt 3)=0$ and two quadratics with real coefficients. One of these will have your original root and its conjugate as solutions, and the other you can solve for the other roots.
