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Byju's Answer
Standard XII
Mathematics
Complex Numbers
Prove that fo...
Question
Prove that for two complex numbers
z
1
,
z
2
,
(
z
1
−
z
2
)
2
=
z
2
1
−
2
z
1
z
2
+
z
2
2
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Solution
Let
z
1
=
x
1
+
i
y
1
And
z
2
=
x
2
+
i
y
2
Consider LHS
(
z
1
−
z
2
)
2
=
(
(
x
1
+
i
y
1
)
−
(
x
2
+
i
y
2
)
)
2
=
(
x
1
+
i
y
1
)
2
+
(
x
2
+
i
y
2
)
2
−
2
(
x
1
+
i
y
1
)
(
x
2
+
i
y
2
)
(
z
1
−
z
2
)
2
=
z
2
1
+
z
2
2
−
2
z
1
z
2
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0
Similar questions
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
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.
lf
z
1
,
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.
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
Q.
If
z
1
and
z
2
are any two complex numbers, then
∣
∣
z
1
+
√
z
1
2
−
z
2
2
∣
∣
+
∣
∣
z
1
−
√
z
1
2
−
z
2
2
∣
∣
is equal to
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