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Byju's Answer
Standard XII
Mathematics
Geometrical Representation of a Complex Number
If Arg z + ...
Question
If
A
r
g
(
z
+
1
z
−
1
)
=
π
6
, then find the locus of z.
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Solution
Given,
z
+
1
z
−
1
=
x
+
i
y
+
1
x
+
i
y
−
1
=
(
x
+
1
)
+
i
y
(
x
−
1
)
+
i
y
×
(
x
−
1
)
−
i
y
(
x
−
1
)
−
i
y
=
(
x
+
1
)
(
x
−
1
)
+
y
2
−
2
i
y
(
x
−
1
)
2
+
y
2
∴
z
=
x
2
−
1
+
y
2
(
x
−
1
)
2
+
y
2
−
i
2
y
(
x
−
1
)
2
+
y
2
θ
=
tan
−
1
(
b
a
)
=
tan
−
1
(
2
y
x
2
+
y
2
−
1
)
tan
π
6
=
1
√
3
1
√
3
=
2
y
x
2
+
y
2
−
1
x
2
+
y
2
−
1
=
2
√
3
y
x
2
+
y
2
+
0
x
−
2
√
3
y
−
1
=
0
g
=
0
,
f
=
−
√
3
,
c
=
−
1
c
e
n
t
e
r
=
(
−
g
,
−
f
)
=
(
0
,
√
3
)
r
a
d
i
u
s
=
√
g
2
+
f
2
−
c
=
√
0
2
+
(
−
√
3
)
2
−
(
−
1
)
=
2
Locus of z is a circle with center
(
0
,
√
3
)
and radius
2
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Standard XII Mathematics
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