Question

If \[{{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},............,{{\alpha }_{n}}\] are the roots of the equation \[{{x}^{n}}+{{p}_{1}}{{x}^{n-1}}+{{p}_{2}}{{x}^{n-2}}+............+{{p}_{n-1}}x+{{p}_{n}}=0\] , \[{{p}_{1}},{{p}_{2}},.......{{p}_{n}}\] being real, prove that
\[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)={{\left( 1-{{p}_{2}}+{{p}_{4}}+...... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}.... \right)}^{2}}\].

Answer 1

Hint:First of all, put the value of \[x=i\] in the equation \[{{x}^{n}}+{{p}_{1}}{{x}^{n-1}}+{{p}_{2}}{{x}^{n-2}}+............+{{p}_{n-1}}x+{{p}_{n}}=0\] . We know that \[{{i}^{2}}=-1\] . Now, substitute \[{{i}^{2}}\] by -1. Then, square the LHS and RHS of the equation \[{{\left\{ i\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right) \right\}}^{2}}={{\left\{ -\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right) \right\}}^{2}}\] and substitute \[{{i}^{2}}\] by -1. Now, get the value of \[{{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}}\] . Take \[{{\alpha }_{1}}={{\alpha }_{2}}={{\alpha }_{3}}=............={{\alpha }_{n}}=i\] and then square it. Again, substitute \[{{i}^{2}}\] by -1 and then, add 1 to it. Now, get the value of \[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)\] . Compare the value of the equation \[{{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}}\] and \[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)\] . Now, solve it further and prove the required equation.

Complete step-by-step solution:
According to the question, it is given that we have an equation
\[{{x}^{n}}+{{p}_{1}}{{x}^{n-1}}+{{p}_{2}}{{x}^{n-2}}+............+{{p}_{n-1}}x+{{p}_{n}}=0\] ……………………………………………..(1)
The roots of the above equation are \[{{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},............,{{\alpha }_{n}}\] ……………………………………(2)
Now, on putting the value of \[x=i\] in equation (1), we get
\[{{i}^{n}}+{{p}_{1}}{{i}^{n-1}}+{{p}_{2}}{{i}^{n-2}}+............+{{p}_{n-1}}i+{{p}_{n}}=0\] ……………………………………..(3)
Now, on taking the term \[{{i}^{n-1}}\] as common in the equation (3), we get
\[\begin{align}
  & \Rightarrow {{i}^{n}}+{{p}_{1}}{{i}^{n-1}}+{{p}_{2}}{{i}^{n-2}}+............+{{p}_{n-1}}i+{{p}_{n}}=0 \\
 & \Rightarrow {{i}^{n-1}}\left( i+{{p}_{1}}+{{p}_{2}}\dfrac{{{i}^{n-2}}}{{{i}^{n-1}}}+{{p}_{3}}\dfrac{{{i}^{n-3}}}{{{i}^{n-1}}}+{{p}_{4}}\dfrac{{{i}^{n-4}}}{{{i}^{n-1}}}+{{p}_{5}}\dfrac{{{i}^{n-5}}}{{{i}^{n-1}}}+............ \right)=0 \\
 & \Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{{{i}^{n-2}}}{{{i}^{n-1}}}+{{p}_{3}}\dfrac{{{i}^{n-3}}}{{{i}^{n-1}}}+{{p}_{4}}\dfrac{{{i}^{n-4}}}{{{i}^{n-1}}}+{{p}_{5}}\dfrac{{{i}^{n-5}}}{{{i}^{n-1}}}+............ \right)=0 \\
 & \Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{1}{{{i}^{\left( n-1 \right)-\left( n-2 \right)}}}+{{p}_{3}}\dfrac{1}{{{i}^{\left( n-1 \right)-\left( n-3 \right)}}}+{{p}_{4}}\dfrac{1}{{{i}^{\left( n-1 \right)-\left( n-4 \right)}}}+{{p}_{5}}\dfrac{1}{{{i}^{\left( n-1 \right)-\left( n-5 \right)}}}+............ \right)=0 \\
\end{align}\]
\[\Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{1}{{{i}^{1}}}+{{p}_{3}}\dfrac{1}{{{i}^{2}}}+{{p}_{4}}\dfrac{1}{{{i}^{3}}}+{{p}_{5}}\dfrac{1}{{{i}^{4}}}+............ \right)=0\] …………………………………..(4)
We know that \[{{i}^{2}}=-1\] …………………………………(5)
Now, from equation (4) and equation (5), we get
\[\begin{align}
  & \Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{1}{i}+{{p}_{3}}\dfrac{1}{\left( -1 \right)}+{{p}_{4}}\dfrac{1}{i\left( -1 \right)}+{{p}_{5}}\dfrac{1}{\left( -1 \right)\left( -1 \right)}+............ \right)=0 \\
 & \Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{1}{i}-{{p}_{3}}-{{p}_{4}}\dfrac{1}{i}+{{p}_{5}}+............ \right)=0 \\
\end{align}\]
\[\Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{i}{{{i}^{2}}}-{{p}_{3}}-{{p}_{4}}\dfrac{i}{{{i}^{2}}}+{{p}_{5}}+............ \right)=0\] …………………………………(6)
Now, on substituting the value of \[{{i}^{2}}=-1\] from equation (5) in equation (6), we get
\[\begin{align}
  & \Rightarrow \left( i+{{p}_{1}}+{{p}_{2}}\dfrac{i}{\left( -1 \right)}-{{p}_{3}}-{{p}_{4}}\dfrac{i}{\left( -1 \right)}+{{p}_{5}}+............ \right)=0 \\
 & \Rightarrow \left( i+{{p}_{1}}-{{p}_{2}}i-{{p}_{3}}+{{p}_{4}}i+{{p}_{5}}+............ \right)=0 \\
 & \Rightarrow i\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)+\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)=0 \\
\end{align}\]
\[\Rightarrow i\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)=-\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)\] ……………………………………..(7)
On squaring the LHS and RHS of equation (7), we get
\[\Rightarrow {{\left\{ i\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right) \right\}}^{2}}={{\left\{ -\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right) \right\}}^{2}}\]
Now, on substituting the value of \[{{i}^{2}}=-1\] from equation (5) in equation (7), we get
\[\begin{align}
  & \Rightarrow \left( -1 \right){{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}={{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}} \\
 & \Rightarrow 0={{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}} \\
\end{align}\]
\[\Rightarrow {{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}}=0\] ………………………………………….(8)
From equation (2), we have \[{{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},............,{{\alpha }_{n}}\] as the roots of the equation \[{{x}^{n}}+{{p}_{1}}{{x}^{n-1}}+{{p}_{2}}{{x}^{n-2}}+............+{{p}_{n-1}}x+{{p}_{n}}=0\] .
Now, taking \[{{\alpha }_{1}}={{\alpha }_{2}}={{\alpha }_{3}}=............={{\alpha }_{n}}=i\] …………………………………(9)
Now, on squaring equation (9), we get
\[{{\alpha }_{1}}^{2}={{\alpha }_{2}}^{2}={{\alpha }_{3}}^{2}=............={{\alpha }_{n}}^{2}={{i}^{2}}\] …………………………………………….(10)
Now, on substituting the value of \[{{i}^{2}}=-1\] from equation (5) in equation (10), we get
\[{{\alpha }_{1}}^{2}={{\alpha }_{2}}^{2}={{\alpha }_{3}}^{2}=............={{\alpha }_{n}}^{2}=-1\] ………………………………………………(11)
Now, on adding 1 in equation (11), we get
\[\left( 1+{{\alpha }_{1}}^{2} \right)=\left( 1+{{\alpha }_{2}}^{2} \right)=\left( 1+{{\alpha }_{3}}^{2} \right)=............=\left( 1+{{\alpha }_{n}}^{2} \right)=-1+1=0\] ………………………………(12)
From equation (12), we have
\[\left( 1+{{\alpha }_{1}}^{2} \right)=0\]
\[\left( 1+{{\alpha }_{2}}^{2} \right)=0\]
\[\left( 1+{{\alpha }_{3}}^{2} \right)=0\]
\[\left( 1+{{\alpha }_{n}}^{2} \right)=0\]
Now, on multiplying, we get
\[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)=0\] ……………………………………..(13)
Now, from equation (8) and equation (13), we have
\[{{\left( 1-{{p}_{2}}+{{p}_{4}}+........... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}+.......... \right)}^{2}}=0\]
\[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)=0\]
The RHS of equation (8) and equation (13) are equal to each other. So, the LHS of equation (8) and equation (13) must be equal to each other.
Therefore, \[\left( 1+{{\alpha }_{1}}^{2} \right)\left( 1+{{\alpha }_{1}}^{2} \right)...\left( 1+{{\alpha }_{1}}^{2} \right)={{\left( 1-{{p}_{2}}+{{p}_{4}}+...... \right)}^{2}}+{{\left( {{p}_{1}}-{{p}_{3}}+{{p}_{5}}.... \right)}^{2}}\] .
Hence, proved.

Note: In this question, one might get confused because the formula for the sum of the roots and products of roots is useless here. So, if we observe the expression which we have to prove, we can see that the signs of the coefficients are changing alternatively. Therefore, whenever we are given an equation and the coefficient of the variable x is changing alternatively, then approach that question by putting the value of the variable x equal to \[i\]