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Standard XII
Mathematics
Properties of Conjugate of a Complex Number
Question
If
|
z
|
=
1
,
|
a
|
≠
1
, and
x
=
z
(
z
−
a
)
(
1
−
z
¯
a
)
, then
x
<
0
x
>
0
0
<
x
<
1
x
≥
1
A
x
≥
1
B
x
<
0
C
x
>
0
D
0
<
x
<
1
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Solution
Verified by Toppr
Given
|
z
|
=
1
,
|
a
|
≠
1
then,
x
=
z
(
z
−
a
)
(
1
−
z
¯
a
)
=
1
(
1
−
a
)
(
1
−
a
)
let
a
=
x
+
i
y
and
¯
a
=
x
−
i
y
=
1
[
1
−
(
x
+
i
y
)
]
[
1
−
(
x
−
i
y
)
]
=
1
1
−
(
x
−
i
y
)
−
(
x
+
i
y
)
+
(
x
+
i
y
)
(
x
−
i
y
)
=
1
1
−
x
+
i
y
−
x
−
i
y
+
x
2
−
(
i
y
)
2
=
1
(
x
−
1
)
2
−
y
2
∴
x
>
0.
Hence, the answer is
x
>
0.
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Similar Questions
Q1
If
|
z
|
=
1
,
|
a
|
≠
1
, and
x
=
z
(
z
−
a
)
(
1
−
z
¯
a
)
, then
View Solution
Q2
If
x
>
0
,
y
>
0
,
z
>
0
,
x
y
+
y
z
+
z
x
<
1
and if
tan
−
1
+
tan
−
1
y
+
tan
−
1
z
=
π
then
x
+
y
+
z
=
View Solution
Q3
If
x
≠
0
,
y
≠
0
,
z
≠
0
and
∣
∣ ∣
∣
1
+
x
1
1
1
+
y
1
+
2
y
1
1
+
z
1
+
z
1
+
3
z
∣
∣ ∣
∣
=
0
then
x
−
1
+
y
−
1
+
z
−
1
is equal to
View Solution
Q4
If
z
+
√
2
|
z
+
1
|
+
i
=
0
and
z
=
x
+
i
y
, then
View Solution
Q5
If
x
,
y
,
z
are different and
∣
∣ ∣ ∣
∣
x
x
2
1
+
x
2
y
y
2
1
+
y
2
z
z
2
1
+
z
2
∣
∣ ∣ ∣
∣
=
0
then prove that
1
+
x
y
z
=
0
.
View Solution