= Solution
Let $t=q+1-\#E(\mathbb F_q)$. The stated characteristic equation for the <Frobenius isogeny> is
$$
\pi^2-[t]\pi+[q]=0.
$$
Take $\alpha,\beta$ to be the complex roots of $T^2-tT+q$. They satisfy $\alpha+\beta=t$ and $\alpha\beta=q$. By the <Hasse bound>, $t^2\le4q$. If the inequality is strict, the roots are complex conjugates and have squared modulus $q$. In the equality case both roots are the same real number $\pm\sqrt q$. Thus in every case
$$
|\alpha|=|\beta|=\sqrt q,\qquad \beta=q/\alpha.
$$
To obtain every extension-field count without assuming distinct roots, set $t_n=\operatorname{tr}(\pi^n)$ and $t_0=2$. The <dual isogeny> satisfies $\widehat\pi\pi=[q]$ and $\pi+\widehat\pi=[t]$. Consequently the sums $\pi^n+\widehat\pi^{\,n}=[t_n]$ satisfy
$$
t_n=t\,t_{n-1}-q\,t_{n-2},\qquad t_0=2,\quad t_1=t.
$$
Alternatively, apply additivity of the <trace of an elliptic-curve endomorphism> directly to the characteristic equation multiplied by $\pi^{n-2}$. The sequence $\alpha^n+\beta^n$ has the same recurrence and initial values, so $t_n=\alpha^n+\beta^n$ by induction. This is the <Frobenius trace recurrence>, valid also at a repeated root.
The fixed points of $\pi^n$ are $E(\mathbb F_{q^n})$, and $1-\pi^n$ is again separable. Since $\deg\pi^n=q^n$, the trace-degree identity gives
$$
\#E(\mathbb F_{q^n})=\deg(1-\pi^n)=1+q^n-t_n
=1+q^n-\alpha^n-\beta^n.
$$
Factoring and using $\beta=q/\alpha$ yields
$$
\boxed{\#E(\mathbb F_{q^n})=(1-\alpha^n)(1-q^n\alpha^{-n}),
\qquad |\alpha|=\sqrt q\quad(n\ge1).}
$$
This is the <elliptic-curve point count over a finite field> formula and the <Riemann hypothesis for an elliptic curve over a finite field>. In particular the powers on both factors are necessary; they are damaged in the converted TeX but present in the original PDF.
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