Let z j - j -1-2-3- - - 7 be the roots of the equation z 7-1- z 7- Then the value of - j -17Re z j - where Re z denotes the real part of z- is

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Standard XII

Mathematics

Relations between Roots and Coefficients : Higher Order Equations

Let z j , j =...

Question

Let $z_{j}, j = 1, 2, 3, . ., 7$ be the roots of the equation $z^{7} = (1 - z)^{7} .$ Then the value of $7 \sum j = 1 R e (z_{j}),$ where $R e (z)$ denotes the real part of $z$ , is

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Solution

If the polynomial equation has real coefficients, then the complex roots occur in conjugate pair.

$z^{7} = (1 - z)^{7} \Rightarrow z^{7} + (z - 1)^{7} = 0 \Rightarrow 2 z^{7} - 7 z^{6} + . . . . = 0 \Rightarrow 7 \sum j = 1 R e (z_{j}) = 7 \sum j = 1 z_{j} = \frac{7}{2} = 3.5$

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Q. Let

z_{j}, j = 1, 2, 3, . ., 7

be the roots of the equation

z^{7} = (1 - z)^{7} .

Then the value of

7 \sum j = 1 R e (z_{j}),

where

R e (z)

denotes the real part of

z

, is

Q. Let

z_{k} (k = 0, 1, 2, . . ., 6)

be the roots of the equation

{(z + 1)}^{7} + z^{7} = 0

, then

6 \sum k = 0 R e (z_{k})

is equal to

Q. Let Let

Z_{k} (k = 0, 1, 2, . . . . . . . . . . . . . . .6)

be the roots of the equation

(z + 1)^{7} + z^{7} = 0,

then

\sum_{k = 0}^{6} R e (z k)

Q. Let

P

be a point denoting a complex number

z

on the complex plane. Let

R e (z)

denotes the real part of

z

and

I m (z)

denotes the imaginary part of

z

. Minimum value of

| z |

such that

| R e (z) | . | I m (z) | = 36

is equal to

Let

z

be any complex number. To factorise the expression of the form (

z^{n} - 1

), we consider the equation

z^{n} = 1

. This equation is solved using De moiver's theorem. Let

1, α_{1}, α_{2} . . . . . . . α_{n - 1}

be the roots of this equation, then

z^{n} - 1 = (z - 1) (z - α_{1}) (z - α_{2}) . . . . . . . (z - α_{n - 1})

. This method can be generalised to factorize any expression of the form

z^{n} - k^{n}

For example,

z^{7} + 1 = Π_{m = 0}^{6} (z - C i S (\frac{2 m π}{7} + \frac{π}{7}))

This can be further simplified as

z^{7} + 1 (z + 1) (z^{2} - 2 z c o s \frac{π}{7} + 1) (z^{2} - 2 z c o s \frac{3 π}{7} + 1) (z^{2} - 2 z c o s \frac{5 π}{7} + 1)

.......

(i)

These factorisations are useful in proving different trigonometric identities e.g. in equation

(i)

if we put

z = i,

then equation

(i)

becomes

(1 - i) = (i + 1) (- 2 i c o s \frac{π}{7}) (- 2 i c o s \frac{3 π}{7}) (- 2 i c o s \frac{5 π}{7})

i.e,

c o s \frac{π}{7} c o s \frac{3 π}{7} c o s \frac{5 π}{7} = - \frac{1}{8}

By using the factorisation for

z^{5} + 1

, the value of

4 s i n \frac{π}{10} c o s \frac{π}{5}

comes out to be :

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Let zj,j=1,2,3,..,7 be the roots of the equation z7=(1−z)7. Then the value of 7∑j=1Re(zj), where Re(z) denotes the real part of z, is

If the polynomial equation has real coefficients, then the complex roots occur in conjugate pair. z7=(1−z)7⇒z7+(z−1)7=0⇒2z7−7z6+....=0⇒7∑j=1Re(zj)=7∑j=1zj=72=3.5

Let $z_{j}, j = 1, 2, 3, . ., 7$ be the roots of the equation $z^{7} = (1 - z)^{7} .$ Then the value of $7 \sum j = 1 R e (z_{j}),$ where $R e (z)$ denotes the real part of $z$ , is

If the polynomial equation has real coefficients, then the complex roots occur in conjugate pair.

$z^{7} = (1 - z)^{7} \Rightarrow z^{7} + (z - 1)^{7} = 0 \Rightarrow 2 z^{7} - 7 z^{6} + . . . . = 0 \Rightarrow 7 \sum j = 1 R e (z_{j}) = 7 \sum j = 1 z_{j} = \frac{7}{2} = 3.5$