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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.

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

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ $|f(i)|=1$ 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. $\endgroup$ – vadim123 Jul 26 '18 at 15:52
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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 $x_1 = x_2 = x_3 = 0$.

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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...

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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 $(1)$: $$(x_1+x_2+x_3-x_1x_2x_3)^2+(x_1x_2+x_1x_2+x_2x_3)^2-2(x_1x_2+x_1x_3+x_2x_3)=0 \iff \\ x_1^2+x_2^2+x_3^2+(x_1x_2)^2+(x_1x_3)^2+(x_2x_3)^2+(x_1x_2x_3)^2=0 \iff \\ x_1=x_2=x_3=0.$$

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