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Byju's Answer
Standard XII
Mathematics
Area Method to Find Condition for Co-Linearity
If complex nu...
Question
If complex number
z
1
,
z
2
and
0
are vertices of equilateral triangle, then
z
2
1
+
z
2
2
−
z
1
z
2
is equal to
A
0
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B
z
1
−
z
2
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C
z
1
+
z
2
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D
1
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Solution
The correct option is
A
0
Given,
z
1
,
z
2
,
0
are vertices of an equilateral triangle.
So, we have
z
2
1
+
z
2
2
+
0
2
=
z
1
z
2
+
z
2
.0
+
0.
z
1
⇒
z
2
1
+
z
2
2
=
z
1
z
2
⇒
z
2
1
+
z
2
2
−
z
1
z
2
=
0
Hence, option A is correct.
Suggest Corrections
0
Similar questions
Q.
If
z
1
and
z
2
are any two complex numbers, then
∣
∣
∣
z
1
+
√
z
2
1
−
z
2
2
∣
∣
∣
+
∣
∣
∣
z
1
−
√
z
2
1
−
z
2
2
∣
∣
∣
is equal to
Q.
Assertion :Let
z
1
,
z
2
,
z
3
be three complex numbers such that
|
3
z
1
+
1
|
=
|
3
z
2
+
1
|
=
|
3
z
3
+
1
|
and
1
+
z
1
+
z
2
+
z
3
=
0
, then
z
1
,
z
2
,
z
3
will represent vertices of an equilateral triangle on the complex plane. Reason:
z
1
,
z
2
,
z
3
represent vertices of an equilateral triangle if
z
2
1
+
z
2
2
+
z
2
3
=
z
1
z
2
+
z
2
z
3
+
z
3
z
1
.
Q.
The three vertices of a triangle are represented by the complex numbers,
0
,
z
1
and
z
2
. If triangle is equilateral, then show that
z
1
2
+
z
2
2
=
z
1
z
2
. Further if
z
0
is circumcentre then prove that
z
1
2
+
z
2
2
=
3
z
0
2
.
Q.
If
z
1
,
z
2
are two complex numbers
(
z
1
≠
z
2
)
satisfying
|
z
2
1
−
z
2
2
|
=
|
¯
¯¯¯
¯
z
2
1
+
¯
¯¯¯
¯
z
2
2
−
2
¯
¯
¯
z
1
¯
¯
¯
z
2
|
, then
Q.
If
z
1
,
z
2
are two different complex numbers satisfying
|
z
2
1
−
z
2
2
|
=
|
¯
¯
¯
z
2
1
+
¯
¯
¯
z
2
2
−
2
¯
¯
¯
z
1
¯
¯
¯
z
2
|
,
then
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