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EAMCET Engineering 2010 Solved Paper
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© examsnet.com
Question : 69
Total: 160
If
z
=
1
+
i
√
3
then
|
Arg
z
|
+
|
Arg
z
|
is equal to
0
π
3
π
2
2
π
3
Validate
Solution:
z
=
1
+
i
√
3
Let
1
+
i
√
3
=
r
(
cos
θ
+
i
s
i
n
θ
)
Then,
r
cos
θ
=
1
. . . (i)
r
sin
θ
=
√
3
. . . (ii)
Eq. (i) + Eq. (ii), we get
r
2
(
s
i
n
2
θ
+
cos
2
θ
)
=
3
+
1
(
∵
s
i
n
2
θ
+
cos
2
θ
=
1
)
r
2
=
4
r
=
2
Eq. (ii)
÷
Eq. (i), we get
⇒
tan
θ
=
√
3
⇒
tan
θ
=
tan
π
3
⇒
θ
=
π
3
So,
z
=
(
1
+
i
√
3
)
=
2
{
cos
π
3
+
i
s
i
n
π
3
}
,
Then
z
=
(
1
−
i
√
3
)
=
2
{
cos
π
3
−
i
s
i
n
π
3
}
=
2
{
cos
(
−
π
3
)
+
i
s
i
n
(
−
π
3
)
}
So
Arg
z
=
π
3
Arg
z
=
−
π
3
|
Arg
z
|
+
|
Arg
z
|
=
|
π
3
|
+
|
−
π
3
|
=
π
3
+
π
3
=
2
π
3
© examsnet.com
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