2
You visited us
2
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Equation of Conics in Complex Form
If the maximu...
Question
If the maximum possible principal argument of the complex number
z
satisfying
|
z
−
4
|
=
R
e
(
z
)
is
k
, then the value of
π
k
is
Open in App
Solution
Let
z
=
x
+
i
y
|
z
−
4
|
=
R
e
(
z
)
⇒
√
(
x
−
4
)
2
+
y
2
=
x
⇒
x
2
−
8
x
+
16
+
y
2
=
x
2
⇒
y
2
=
8
(
x
−
2
)
This represents a parabola whose directrix is
x
=
0
arg
(
z
)
is maximum possible when
The line drawn from origin is tangent to the parabola
As
x
=
0
is directrix, so angle between the pair of tangents is
π
2
Therefore, the maximum possible value is
arg
(
z
)
=
π
4
∴
π
k
=
4
Suggest Corrections
0
Similar questions
Q.
If the maximum possible principal argument of the complex number
z
satisfying
|
z
−
4
|
=
R
e
(
z
)
is
k
, then the value of
π
k
is
Q.
The greatest positive argument of complex number satisfying
|
z
−
4
|
=
R
e
(
z
)
is
Q.
Let
S
be the set of all complex numbers
z
satisfying
|
z
−
2
+
i
|
≥
√
5
. If the complex number
z
0
is such that
1
|
z
0
−
1
|
is the maximum of the set
{
1
|
z
−
1
|
:
z
∈
S
}
,then the principal argument of
4
−
z
0
−
¯
¯¯¯
¯
z
0
z
−
¯
¯¯¯
¯
z
0
+
2
i
is
Q.
If
k
+
|
k
+
z
2
|
=
|
z
|
2
,
(
k
∈
R
−
)
, then possible argument of
z
is
Q.
If
z
is a complex number satisfying
|
z
3
+
z
−
3
|
≤
2
, then the maximum possible value of
|
z
+
z
−
1
|
is
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Explore more
Equation of Conics in Complex Form
Standard XII Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
AI Tutor
Textbooks
Question Papers
Install app