Question

10. The maximum number of zeroes of a polynomial of degree 'n' is
(a) n+1 (b) n-1 (c) n
(d) none of these.​

Answer 1

Answer:

⇒ Let (1+x)

n

=1+c

1

x+c

2

x

2

+...

(1+ix)

n

=1+ic

1

x−c

2

x

2

+ic

3

x

3

+c

4

x

4

+ic

5

x

5

−....

(1−ix)

n

=1−ic

1

x−c

2

x

2

+ic

3

x

3

+c

4

x

4

−ic

5

x

5

−....

∴2ix(c

1

−c

3

x

2

+ic

5

x

4

−....)=(1+ix)

n

−(1−ix)

n

Put x

2

=3, so that x=

3

, and let S

1

denote the value of the first series;also as usual

Let w,w

2

be the imaginary cube roots of unity;

so that w=

2

−1+

−3

;w

2

=

2

−1−

−3

We have

2i

3

S

1

=(1+

−3

)

n

−(1−

−3

)

n

=(−2w

2

)

n

−(−2w)

n

=2

n

−2

n

=0

when n is a multiple of 6, for then

(−w)

n

=1,(−w

2

)

n

=1

Put x

2

=

3

1

and let S

2

denote the sum of the series, them;-

3

2i

S

2

=(1+

3

−1

)

n

−(1−

3

−1

)

n

=(

−3

−3

−1

)

n

−(

−3

−3

+1

)

n

=(

−3

2w

)

n

−(

−3

−2w

2

)

n

=0