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UPSC NDA 2 2016 Mathematics Solved Paper
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© examsnet.com
Question : 9
Total: 120
If
z
=
(
√
3
2
+
i
2
)
107
+
(
√
3
2
−
i
2
)
107
then what is the imaginary part of z equal to?
[NDA II 2016]
0
1
2
√
3
2
1
Validate
Solution:
z
=
[
√
3
2
+
i
2
]
107
+
[
√
3
2
−
i
2
]
107
Q
cos
π
6
=
√
3
2
&
sin
π
6
=
1
2
⇒
z
=
[
cos
π
6
+
i
sin
π
6
]
107
+
[
cos
π
6
−
i
sin
π
6
]
107
Also
(
cos
θ
+
i
sin
θ
)
n
=
cos
n
θ
+
i
sin
n
θ
⇒
z
=
cos
107
π
6
+
i
sin
107
π
6
+
cos
107
π
6
−
i
sin
107
π
16
⇒
Im
(
z
)
=
0
© examsnet.com
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