Name: Geometric View of Complex Number
Uploaded: 2019-07-19
Duration: 10 min 19 s
Description: Complex numbers extend the concept of the one-dimensional number line to the two-dimensional complex plane by using the horizontal axis. Watch the video to l...

Question

If

α_{1}, α_{2}, α_{3}, \dots, α_{n}

are roots of the equation

x^{n} + p_{1} x^{n - 1} + p_{2} x^{n - 2} + \dots + p_{n - 1} x + p_{n} = 0

, then prove that

{(1 - p_{2} + p_{4} + \dots)}^{2} + {(p_{1} - p_{3} + p_{5} \dots)}^{2} = (1 + α_{1}^{2}) (1 + α_{2}^{2}) (1 + α_{3}^{2}) \dots (1 + α_{n}^{2})

.

Accepted Answer

Given that, $α_{1}, α_{2}, α_{3}, \dots, α_{n}$ are roots of the equation $x^{n} + p_{1} x^{n - 1} + p_{2} x^{n - 2} + \dots + p_{n - 1} x + p_{n} = 0$

We know that, $x^{n} + p_{1} x^{n - 1} + p_{2} x^{n - 2} + \dots + p^{n} = (x - α_{1}) (x - α_{2}) \dots (x - α_{n})$ is an identity.

Put $x = i$ on both sides, we get

$\Rightarrow i^{n} + p_{1} i^{n - 1} + p_{2} i^{n - 2} + \dots + p_{n} = (i - α_{1}) (i - α_{2}) \dots (i - α_{n})$

Taking $i^{n}$ common from LHS and $i$ form RHS, we get

$\Rightarrow i^{n} [1 + \frac{p_{1}}{i} + \frac{p_{2}}{i^{2}} + \dots + \frac{p_{n}}{i^{n}}] = i (1 - \frac{α_{1}}{i}) i (1 - \frac{α_{2}}{i}) \dots i (1 - \frac{α_{n}}{i})$

$\Rightarrow i^{n} [1 + \frac{p_{1}}{i} + \frac{p_{2}}{i^{2}} + \dots + \frac{p_{n}}{i^{n}}] = i^{n} (1 - \frac{α_{1}}{i}) (1 - \frac{α_{2}}{i}) \dots (1 - \frac{α_{n}}{i})$

$\Rightarrow [1 + \frac{p_{1}}{i} + \frac{p_{2}}{i^{2}} + \dots + \frac{p_{n}}{i^{n}}] = (1 - \frac{α_{1}}{i}) (1 - \frac{α_{2}}{i}) \dots (1 - \frac{α_{n}}{i})$

$\Rightarrow 1 - i p_{1} - p_{2} + i p_{3} + p_{4} + \dots = (1 + i α_{1}) (1 + i α_{2}) \dots (1 + i α_{n}) [∵ \frac{1}{i} = - i]$

$\Rightarrow (1 - p_{2} + p_{4} - . . .) - i (p_{1} - p_{3} + p_{5} \dots) = (1 + i α_{1}) (1 + i α_{2}) \dots (1 + i α_{n})$

Apply modulus on both sides, we get

$\Rightarrow |(1 - p_{2} + p_{4} - . . .) - i (p_{1} - p_{3} + p_{5} \dots)| = |(1 + i α_{1}) (1 + i α_{2}) \dots (1 + i α_{n})|$

$\Rightarrow |(1 - p_{2} + p_{4} - . . .) - i (p_{1} - p_{3} + p_{5} \dots)| = |(1 + i α_{1})| |(1 + i α_{2})| \dots |(1 + i α_{n})| [∵ |z_{1} z_{2} . . . z_{n}| = |z_{1}| |z_{2}| . . . |z_{n}|]$

$\Rightarrow \sqrt{{(1 - p_{2} + p_{4} + \dots)}^{2} + {(p_{1} - p_{3} + p_{5} \dots)}^{2}} = \sqrt{(1 + α_{1}^{2})} \sqrt{(1 + α_{2}^{2})} \sqrt{(1 + α_{3}^{2})} \dots \sqrt{(1 + α_{n}^{2})} [∵ |x + i y| = \sqrt{x^{2} + y^{2}}]$

$\Rightarrow {(1 - p_{2} + p_{4} + \dots)}^{2} + {(p_{1} - p_{3} + p_{5} \dots)}^{2} = (1 + α_{1}^{2}) (1 + α_{2}^{2}) (1 + α_{3}^{2}) \dots (1 + α_{n}^{2})$ (Squaring on both sides)

Hence, proved.

M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 7: EXERCISE 5.7

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 7: EXERCISE 5.7

Questions from M L Aggarwal Solutions for Chapter: Complex Numbers, Exercise 7: EXERCISE 5.7 with Hints & Solutions

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