complex numbers two problems

Question

1) If $z=\cos\alpha+i\sin\alpha$ for $\alpha \in[0, 2\pi]$ then find $\alpha$ for $z^2+z$

I transform to this moment $\displaystyle z^2+z=2\cos\frac{\alpha}{2}(\cos\frac{3\alpha}{2}+i\sin\frac{3\alpha}{2})$ and I'm not sure if can now write that $\displaystyle \alpha \in[0, \frac{4}{3}]$ ? Because $\displaystyle 0 \le\frac{3}{2}\alpha \le2\pi$ ???

2) I have $\displaystyle |z+\frac{1}{z}|=2$ and I can't find solution when taking $z=a+bi$ we have $a=\sqrt{2} \vee a=-\sqrt{2}$ and $b=1 \vee b=-1$ I have the rest ones but can't figure out how to get these

Answer 1

let $z=a+bi$ then we get $z+\frac{1}{z}=a+bi+\frac{a-bi}{a^2+b^2}=a\left(\frac{a^2+b^2+1}{a^2+b^2}\right)+ib\left(\frac{a^2+b^2-1}{a^2+b^2}\right)$ therefore we have $\frac{1}{a^2+b^2}\sqrt{(a(a^2+b^2+1))^2+(b(a^2+b^2-1))^2}=2$ I will post the solution soon squaring this and factorizing we get ${\frac { \left( {a}^{2}+{b}^{2}+2\,b-1 \right) \left( {a}^{2}+{b}^{2} -2\,b-1 \right) }{{a}^{2}+{b}^{2}}} =0$

Answer 2

2.)

\begin{align} 2^2 &= \left|z + \frac{1}{z}\right|^2 \\ 4 &= \left(z + \frac{1}{z}\right)\left(z + \frac{1}{z}\right)^* \\ 4 &= zz^* + 2\Re\left(\frac{z}{z^*}\right) + \frac{1}{zz^*}\\ 4 &= a^2 + b^2 + 2\Re\left(\frac{a+ib}{a-ib}\right) + \frac{1}{a^2 + b^2} \\ 4 &= a^2 + b^2 + 2\frac{a^2 - b^2}{a^2+b^2} + \frac{1}{a^2 + b^2} \\ 4 &= (a^2 + b^2)^2 + 2(a^2 - b^2) + 1 \\ 4 &= (A + B)^2 + 2(A-B) + 1 \end{align}

where $A := a^2$ and $B := b^2$. Now try to find all solutions $A,B$ of this monster :-) It's a quadratic equation, so use the well-known solving formula to find a representation $B = f(A)$ and then obtain $a,b$. It's an ugly problem for sure with that approach.