Question

If a,b.c...k are the roots of the equation
$x^n+p_1{x}^{n-1}+p_2{x}^{n-2}.....+P_{n-1}x+P_n=0(p_1,p_2,p_3....arereal),thenprovethat(1+a^2)(1+b^2)...(1+k^2)=(1-p_2+p_4-....{)}^2+(p_1-p_3+p_5-..{)}^2$

Answer 1

Since a, b, c....k are the roots of the given equation , we have the identity
$x^n+p_1{x}^{n-1}+p_2{x}^{n-2}.....+P_{n-1}x+P_n$
= (x - a)(x - b)(x - c)......(x - k). ...(1)
in the identity (1) put x = i
Then $i^n+p_1{i}^{n-1}+p_2{i}^{n-1}+...p_{n-1}i+p_n$
= (i - a)(i - b)(i - c)......(i - k).
or $i^n[1+p_1{i}^{-1}+p_2{i}^{-2}+p_3{i}^{-3}+p_4{i}^{-4}....p_{n-1}{i}^{-(n-1)}+p_n{i}^{-n}]$
= (i - a)(i - b)(i - c)......(i - k)
But $i^{-1}=\frac{1}{i}-i,i^{-2}=\frac{1}{i^2}-1,i^{-3}=\frac{1}{i^3}=i,i^{-4}=\frac{1}{i^4}=1etc.$
$\therefore$ The above identity may be written as
$i^n\left[\right(1-p_2+p_4-..)-i(p_1-p_3+p_5-..\left)\right]$
= $(-1{)}^n$ (a - i)(b - i)(c - i).....(k - i)......(2)
Similarly putting x = - i in (1), we shall obtain
$(-1{)}^n[(1-p_2+p_4-..)-i(p_1-p_3+p_5-..)]$
= $(-1{)}^n$ (a - i)(b - i)(c - i).....(k - i)......(3)
Multiplying (2) and (3), we get
$(-1{)}^n.i^{2n}[(1-p_2+p_4-..{)}^2-i(p_1-p_3+p_5-..{)}^2]$
$=(-1{)}^{2n}(a^2-i^2\left)\right(b^2-i^2\left)\right(c^2-i^2)....(k^2-i^2)$
$\therefore (-1{)}^n{i}^{2n}=(-1{)}^n(-1{)}^n=(-1{)}^{2n}=1,$
this gives
$(1-p_2+p_4-..{)}^2+(p_1-p_3+p_5-..{)}^2$
$=(a^2+1)(b^2+1)(c^2+1).....(k^2+1)$