# Prove that all three roots of $f(x)=0$ are zero. Also prove that $a+b+c=0$. [closed]

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.

## closed as off-topic by Namaste, Isaac Browne, Lord Shark the Unknown, Xander Henderson, Mostafa AyazJul 27 '18 at 17:05

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Isaac Browne, Lord Shark the Unknown, Xander Henderson, Mostafa Ayaz
If this question can be reworded to fit the rules in the help center, please edit the question.

• $|f(i)|=1$ means that $f(i)$ has unit complex norm. $f(i)$ could be $1,-1, i, -i$, or others. It does not mean that $f(i)$ takes on multiple values, it can only be a single value, as $f(x)$ is a polynomial. – vadim123 Jul 26 '18 at 15:52

Let $x_1, x_2, x_3$ be the roots. We have $f(x) = (x - x_1)(x - x_2)(x-x_3)$.
so $x_1 = x_2 = x_3 = 0$.
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...
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 $(1)$: $$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \iff \\ x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \iff \\ x_1=x_2=x_3=0.$$