Finding $f(1)$ of general biquadratic polynomial whose $|f(\iota)|$ is given.

Question

Question: Let $f(x)=x^4+ax^3+bx^2+cx+d$ with real coefficients and real roots. If $|f(\iota)|=1$.
Find the value of $a+b+c+d$.

Since $a+b+c+d=f(1)-1$. Finding $f(1)$ would be quite helpful and it is the shortest way to find the answer (purely my reasoning). I got a feeling that $f(1)=1$ due to some reason. Maybe because $|f(\iota)|=1$ and that $f(1)=|f(1)|$ if $f(1)$ will be positive. It doesn't make sense though.

Here is the obvious approach which I don't like and also am stuck at a point.

$f(\iota)=(1-b+d)+(c-a)\iota$
So,$|f(\iota)|=\sqrt{(1-b+d)^2+(c-a)^2}$
Which gives, $a^2+b^2+c^2+d^2-2b+2d-2bd-2ac=0$
This equation doesn't add anything.

I don't like these usual approaches to the question.
I first want you to consider the initial short approach. If this question is solvable like that then give me some hint otherwise tell me how to proceed further from the equation I produced in $a,b,c,d$

Well $\iota=\sqrt{-1}$. I have read it in quite a no. of books. Our teachers too prefer to use $\iota$ over $i$ — Love Invariants, Dec 26 '18 at 16:20
Maybe thats because of difference in countries. Here everyone uses $\iota$ — Love Invariants, Dec 26 '18 at 16:22
Have a look at this site, too (many countries, not just one). Or this question perhaps. I haven't seen iota, but always $i$. See also here. But indeed, why not iota.... — Dietrich Burde, Dec 26 '18 at 16:24
Its ok. Seems like all other countries use $i$ except for India. I made it clear that I mean $\sqrt{-1}$ — Love Invariants, Dec 26 '18 at 16:31

score 4 · Accepted Answer · answered Dec 26 '18 at 16:47

4

We have $f(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$, $x_i$ real. $$1=|f(i)|^2=|(i-x_1)(i-x_2)(i-x_3)(i-x_4)|^2=(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)\geq 1$$

Equality holds, so $x_1=x_2=x_3=x_4=0$, so $a=b=c=d=0$, and their sum is $0$.

answered Dec 26 '18 at 16:47

user629535

56

Great solution! – nonuser Dec 26 '18 at 16:53
1

Great. I don't know that "Equality holds" is really the best explanation for the punchline; the point is that each $x_j^2+1$ is $\ge 1$, hence equality holds in each inequality $x_j^1+1\ge1$. – David C. Ullrich Dec 26 '18 at 17:00

Finding $f(1)$ of general biquadratic polynomial whose $|f(\iota)|$ is given.

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