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Joint Entrance Examination
Mathematics
Complex Numbers
Question 441
Question 441 of 455
Medium
MCQ
For all \[z \in C\] on the curve
C
1
:
∣
z
∣
=
4
C_{1}:|z|=4
C
1
:
∣
z
∣
=
4
, let the locus of the point \[z+\frac{1}{z}\] be the curve \[\mathrm{C}_{2}\]. Then :
A
the curves \[C_{1}\] and \[C_{2}\] intersect at 4 points
B
the curve \[C_{2}\] lies inside \[C_{1}\]
C
the curve \[C_{1}\] lies inside \[C_{2}\]
D
the curves \[C_{1}\] and \[C_{2}\] intersect at 2 points
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Official solution
Let
w
=
z
+
1
z
=
4
e
i
θ
+
1
4
e
−
i
θ
\mathrm{w}=\mathrm{z}+\frac{1}{\mathrm{z}}=4 \mathrm{e}^{\mathrm{i} \theta}+\frac{1}{4} \mathrm{e}^{-\mathrm{i} \theta}
w
=
z
+
z
1
=
4
e
i
θ
+
4
1
e
−
i
θ
⇒
w
=
17
4
cos
θ
+
i
15
4
sin
θ
\Rightarrow \mathrm{w}=\frac{17}{4} \cos \theta+\mathrm{i} \frac{15}{4} \sin \theta
⇒
w
=
4
17
cos
θ
+
i
4
15
sin
θ
So locus of
w
w
w
is ellipse
x
2
(
17
4
)
2
+
y
2
(
15
4
)
2
=
1
\frac{x^{2}}{\left(\frac{17}{4}\right)^{2}}+\frac{y^{2}}{\left(\frac{15}{4}\right)^{2}}=1
(
4
17
)
2
x
2
+
(
4
15
)
2
y
2
=
1
Locus of
z
\mathrm{z}
z
is circle
x
2
+
y
2
=
16
\mathrm{x}^{2}+\mathrm{y}^{2}=16
x
2
+
y
2
=
16
So intersect at 4 points.
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