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AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper
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© examsnet.com
Question : 10
Total: 160
If z = cos 6° + i sin 6° , then
20
Σ
n
=
1
Im
(
z
2
n
−
1
)
=
0
- 1
−
3
4
s
i
n
6
°
3
4
s
i
n
6
°
Validate
Solution:
20
∑
n
=
1
z
2
n
−
1
=
z
+
z
3
+
z
5
+
.
.
.
+
z
39
Now
,
z
=
cos
6
∘
+
i
sin
6
∘
=
e
i
6
∘
Substitute the value in the series,
20
∑
n
=
1
z
2
n
−
1
=
e
i
6
∘
+
e
i
18
∘
+
e
i
30
∘
+
.
.
.
+
e
i
(
39
×
6
∘
)
=
e
i
6
∘
(
1
+
e
i
12
+
e
i
24
+
.
.
.
.
.
+
e
i
228
)
=
e
i
6
∘
[
e
i
240
−
1
e
i
12
−
1
]
Simplify the above equation,
20
∑
n
=
1
Im
(
z
2
n
−
1
)
=
e
i
6
∘
[
e
i
240
−
1
e
i
12
−
1
]
=
e
i
6
∘
[
−
1
2
−
√
3
2
i
−
1
]
e
i
6
∘
(
2
i
sin
6
∘
)
=
[
−
3
2
−
√
3
2
i
]
2
i
sin
6
∘
=
−
√
3
4
sin
6
∘
+
√
3
4
sin
6
∘
i
Thus
,
20
∑
n
=
1
Im
(
z
2
n
−
1
)
=
3
4
sin
6
∘
© examsnet.com
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