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JEE Main 9 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 4
Total: 90
Let
z
be a complex number such that the real part of
z
−
2
i
z
+
2
i
is zero. Then, the maximum value of
|
z
−
(
6
+
8
i
)
|
is equal to :
[9 Apr 2024 Shift 2]
12
∞
10
8
Validate
Solution:
👈: Video Solution
z
−
2
i
z
+
2
i
+
z
+
2
i
z
−
2
i
=
0
z
z
−
2
i
z
−
2
i
z
+
4
(
−
1
)
+
z
z
+
2
z
i
+
2
z
i
+
4
(
−
1
)
=
0
⇒
2
|
z
|
2
=
8
⇒
|
z
|
=
2
∣
|
z
−
(
6
+
8
i
)
|
maximum
=
10
+
2
=
12
© examsnet.com
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