If $ w_1 $and $ w_2 $ are the roots of $w^2-2sw+t=0$, then $|w_1|=|w_2|\iff 0Let $s$ and $t$ be two complex numbers not equal $0$, $ w_1 $and $ w_2 $ are solutions to this equation  $$w^2-2sw+t=0$$
How can prove this equivalence ?
$$|w_1|=|w_2|\iff 0<\dfrac{s^2}{t}\le1$$.
I want a clever ! solution !
So, here's what I've prove so far: $\dfrac{s^2}{t}\in\Bbb R $ and $|\dfrac{s^2}{t}|\le1$.(in the forward direction)
but don't know what do do next. There should be better methods, please suggest some of them.
 A: Sketch of a proof:
$w^2-2sw+t=0$
$\Rightarrow w_{1,2} = s \pm \sqrt{s^2 - t}$
For $|w_1| = |w_2|$ we require
$s = 0$ or
$s^2 - t = 0$ or
$\sqrt{s^2-t}$ perpendicular to $s$.
i.e. $\sqrt{1-\dfrac{t}{s^2}}$ perpendicular to $1$.
i.e. $1-\dfrac{t}{s^2}$ is a negative number.
$\Rightarrow 0 < \dfrac{s^2}{t} < 1$
A: One part of your question: $0<\dfrac{s^2}{t}\le1\Rightarrow|w_1|=|w_2|$. Since $\dfrac{s^2}{t}\le1$ then the discriminant is negative hence you have two complex solutions, so the must be complex conjugate, so you have $|w_1|=|w_2|$.
For the second part: $|w_1|=|w_2|\Rightarrow 0<\dfrac{s^2}{t}\le1$.
Let's assume that $|w_1|=|w_2|$. For a quadratic equation that can be possible if you have two distinct complex conjugate roots or one repeated root. The first case takes place when you have $\dfrac{s^2}{t}<1$ and the second only when $\dfrac{s^2}{t}=1$.
UPDATE
Since $s$ and $t$ are complex (I've overlooked it at first) the things are getting a bit more complex :).
Then will use exponential form and Vieta's formulas. So
$$s=|s|e^{i\arg(s)}, \ t=|t|e^{i\arg(t)}, \ w_1=|w_1|e^{i\arg(w_1)} \ \text{and} \ \ w_2=|w_2|e^{i\arg(w_2)}$$
$$w_1w_2=t \ \text{and} \ w_1+w_2=2s $$
Then (since $|w_1|=|w_2|=|w|$)
$$w_1w_2=|w_1||w_2|e^{i(\arg(w_1)+\arg(w_2))}=|t|e^{i\arg(t)}\Rightarrow |t|=|w|^2 \text{and} \ \arg(w_1)+\arg(w_2)=\arg(t) $$
$$2s=2|s|e^{i\arg(s)}=|w_1|e^{i\arg(w_1)}+|w_2|e^{i\arg(w_2)}=|w|e^{i\frac{(\arg(w_1)+\arg(w_2))}{2}}\left(e^{i\frac{(\arg(w_1)-\arg(w_2))}{2}}+e^{-i\frac{(\arg(w_1)-\arg(w_2))}{2}}\right)=2|w|e^{i\frac{(\arg(w_1)+\arg(w_2))}{2}}\cos\left(\frac{\arg(w_1)-\arg(w_2)}{2}\right)$$
Then
$$\frac{s^2}{t}=\frac{\left(|w|e^{i\frac{(\arg(w_1)+\arg(w_2))}{2}}\cos\left(\frac{\arg(w_1)-\arg(w_2)}{2}\right)\right)^2}{|w|^2e^{i(\arg(w_1)+\arg(w_2))}}=\cos^2\left(\frac{\arg(w_1)-\arg(w_2)}{2}\right)$$
And hence $0<\frac{s^2}{t}\leq 1$.
A: Note we cannot have $w_1 = 0$ or $w_2 = 0$.  Given that $w_1, w_2$ are solutions to $f(w) = w^2 - 2sw + t = 0$, then we note that $$s = \frac{w_1 + w_2}{2}, \quad t = w_1 w_2.$$  Therefore, $$\frac{s^2}{t} = \frac{(w_1 + w_2)^2}{4 w_1 w_2} = \frac{\left( u + u^{-1} \right)^2}{4},$$ where $u = \sqrt{w_1/w_2}$.  If $|w_1| = |w_2|$, then clearly $|u| = 1$, hence $u+u^{-1}$ is a pure real number with zero imaginary part.  It also follows that $s^2/t > 0$ since the RHS is a square.  But since $|u| = |u^{-1}| = 1$, we have by the triangle inequality $$|u+u^{-1}| \le |u| + |u^{-1}| = 2,$$ hence $|s^2/t| \le 2^2/4 = 1$, and the forward direction is proved.
I will give the reverse direction some thought when I have more time.

Addendum.  On second thought, given the lack of work shown in the question and the implied demand for a "simple" proof, I see no reason to furnish a complete proof.  One direction satisfying the asker's criteria is more than generous as it is.
A: A geometric framing of this: First, note that we may assume $|w_1|=1$ without loss of generality (why?). Then the second root $t\in\mathbb{C}$ must satisfy 
\begin{align}
(w-z)&(w-1)=w^2-(1+z)w+t\\
&\implies s=\frac{1+t}{2}\\
&\implies \frac{s^2}{t}=\frac{1}{4}(1+t)(1+t^{-1})
\end{align}
Since $t^{-1}=t^*$ iff $|t|=1$, this maps $t$ on the unit circle to $\left|\dfrac{1+t}{2}\right|^2$. But the image of the unit circle under $\dfrac{1+t}{2}$ is just a circle of radius $1/2$ centered at $1/2$, which passes through the origin and is never is farther than $1$ from the origin. Hence the unit circle is mapped to the interval $[0,1]$---that is, $0\leq s^2/t \leq 1$ if $|t|=1$. (To get the 'only if', note that $t=0$ and $t=\infty$ are both mapped to $s^2/t=\infty$; since the map $t\mapsto \frac{1}{4}(1+t)(1+t^{-1})$ is locally 1-to-1, it follows(?) that the images of $|t|<1$ and $|t|>1$ are both $\mathbb{C}-[0,1]$).
