<s>[INST] Matrix from eigenvalues $A=\begin{bmatrix}
a & b\\ 
b & c
\end{bmatrix}$
Find any $(a,b,c) \in \mathbb{C}^3$ , $a \neq 0$ , $b \neq 0$ , $c \neq 0$ ,  so that eigenvalues of A are $\lambda_1=\lambda_2=1$

$det(A-\lambda I)=\begin{bmatrix}
a-\lambda & b\\ 
b & c- \lambda
\end{bmatrix}=(a-\lambda)(c-\lambda)-b^2=0$
$\Longrightarrow$
$\lambda^2-\lambda(a+c)+ac-b^2=0$
$\lambda=\frac{a+c \pm \sqrt{a^2+4b^2-2ac+c^2}}{2}$
$\Longrightarrow$
$a^2+4b^2-2ac+c^2=0$ and $\frac{a+c}{2}=1$
(discriminant is 0 because I want a double root, and (a+c)/2=1 because I want it to be 1)

but I can't find any solution, is there something wrong?

$\Longrightarrow$
$c=2-a$ 
and 
$4a^2-8a+4b^2+4=0$
$a= 1 \pm b \cdot i $
Whatever value I give to a, it won't work...
 [/INST] Your work looks OK up to $4a^2-8a+4b^2+4=0$.
This reduces to $a^2-2a+b^2+1=0$, and you can try a few values for $a$ and $b$. I tried $a=3$, which yielded $c=-1$ and $b=2i$. The resulting matrix has the eigenvalues you want.
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