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.
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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
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