Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$. 
Let $f(x)=x^3+ax^2+bx+c$ be a cubic polynomial with real coefficients and all real roots, also $|f(i)|=1$ where $i=\sqrt{-1}$. Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$.


As $f(i)=-i-a+ib+c=1$ and $f(i)=-i-a+ib+c=-1$
I don't know how to solve further.
 A: Let $x_1, x_2, x_3$ be the roots. We have $f(x) = (x - x_1)(x - x_2)(x-x_3)$.
Hence
\begin{align}
1 &= |f(i)|^2 \\
&= f(i)\overline{f(i)} \\
&= (i - x_1)(i - x_2)(i - x_3)(-i - x_1)(-i - x_2)(-i - x_3) \\
&= (x_1^2 + 1)(x_2^2 + 1)(x_3^2 + 1)
\end{align}
so $x_1 = x_2 = x_3 = 0$.
A: Hint: Let the three real roots of $f(x)$ be $r,s,t$ (not necessarily distinct).  Then, we may write $$f(x)=(x-r)(x-s)(x-t)$$
Expand this, and set equal to $x^3+ax^2+bx+c$, and continue...
A: Alternatively, note that for $z=a+bi\in \mathbb Z, \bar{z}=a-bi \in \mathbb Z$, the norm is:
$$|z|=\sqrt{z\cdot \bar{z}}=\sqrt{a^2+b^2}.$$
So:
$$|f(i)|=|c-a+(b-1)i|=\sqrt{(c-a)^2+(b-1)^2}=1 \Rightarrow (c-a)^2+b^2-2b=0 \tag{1}$$
Let $x_1,x_2,x_3$ be the roots of $f(x)=0 \iff x^3+ax^2+bx+c=0.$ 
By the Vieta's formulas:
$$\begin{align}\begin{cases}
x_1+x_2+x_3&=-a\\ 
x_1x_2+x_1x_3+x_2x_3&=b\\
x_1x_2x_3&=-c\end{cases} \tag{2}\end{align}$$
Plug $(2)$ to $(1)$:
$$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \iff \\
x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \iff \\
x_1=x_2=x_3=0.$$
