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Byju's Answer
Standard XII
Mathematics
Properties of nth Root of a Complex Number
∑k=16sin2π k7...
Question
6
∑
k
=
1
(
sin
2
π
k
7
−
i
cos
2
π
k
7
)
equals
A
1
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B
i
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C
-i
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D
-1
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Solution
The correct option is
B
i
Solution:
6
∑
k
=
1
(
sin
2
π
k
7
−
i
cos
2
π
k
7
)
=
6
∑
k
=
1
−
i
(
cos
2
π
k
7
+
i
sin
2
π
k
7
)
=
−
i
6
∑
k
=
1
e
i
2
k
π
7
=
−
i
⎡
⎢
⎣
e
i
2
π
7
+
e
i
4
π
7
+
e
i
6
π
7
+
e
i
8
π
7
+
e
i
10
π
7
+
e
i
12
π
7
⎤
⎥
⎦
=
−
i
⎡
⎢ ⎢ ⎢
⎣
e
i
2
π
7
⎛
⎜ ⎜ ⎜
⎝
1
−
e
i
12
π
7
1
−
e
i
2
π
7
⎞
⎟ ⎟ ⎟
⎠
⎤
⎥ ⎥ ⎥
⎦
=
−
i
⎡
⎢ ⎢ ⎢
⎣
⎛
⎜ ⎜ ⎜
⎝
e
i
2
π
7
−
e
i
14
π
7
1
−
e
i
2
π
7
⎞
⎟ ⎟ ⎟
⎠
⎤
⎥ ⎥ ⎥
⎦
=
−
i
⎡
⎢ ⎢ ⎢
⎣
⎛
⎜ ⎜ ⎜
⎝
e
i
2
π
7
−
1
1
−
e
i
2
π
7
⎞
⎟ ⎟ ⎟
⎠
⎤
⎥ ⎥ ⎥
⎦
∵
(
e
i
14
π
7
=
1
)
=
−
i
×
(
−
1
)
=
i
Hence, A is the correct option.
Suggest Corrections
0
Similar questions
Q.
The value of
S
=
∑
6
k
=
1
(
sin
2
π
k
7
−
i
cos
2
π
k
7
)
?
Q.
The value of
100
∑
k
=
1
[
sin
(
2
π
k
101
)
−
i
cos
(
2
π
k
101
)
]
is
Q.
The sum
4
n
+
1
∑
m
=
1
[
m
+
1
∑
k
=
1
m
(
sin
(
2
π
k
m
)
−
i
cos
(
2
π
k
m
)
)
]
m
is
Q.
The value of
10
∑
k
=
1
(
sin
2
π
k
11
−
i
cos
2
π
k
11
)
is
Q.
If
S
k
=
sin
(
2
π
k
n
+
1
)
−
i
cos
(
2
π
k
n
+
1
)
and
P
k
=
cos
(
π
2
k
)
+
i
sin
(
π
2
k
)
,
Let
m
=
∣
∣
∣
n
∑
k
=
1
S
k
+
∞
∏
k
=
1
P
k
∣
∣
∣
, then the value of
2
m
2
is
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