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TS EAMCET 04 May 2019 Shift 2 Solved Paper
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© examsnet.com
Question : 9
Total: 160
If p is a complex number whose modulus is one. then the equation
(
1
+
i
z
1
−
i
z
)
4
= p has
real and equal roots
real and distinct roots
two real and two complex roots
all complex roots
Validate
Solution:
Consider the complex number
(
1
+
i
z
1
−
i
z
)
4
=
p
The above complex number is solved as
(
1
+
i
z
1
−
i
z
)
4
=
p
(
i
−
z
i
+
z
)
4
=
p
|
z
−
i
z
+
i
|
4
=
|
p
|
|
z
−
i
z
+
i
|
4
=
1
Thus,
|
z
−
i
|
4
=
|
z
+
i
|
4
|
z
−
i
|
=
|
z
+
i
|
Therefore,
z
lies on perpendicular bisector of
i
and
−
i
So
z
lies on
y
axis.
Therefore,
z
has all complex roots.
© examsnet.com
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