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Joint Entrance Examination
Mathematics
Complex Numbers
Question 164
Question 164 of 455
Medium
MCQ
If
z
=
x
+
i
y
z=x+iy
z
=
x
+
i
y
satisfies
∣
z
∣
−
2
=
0
|z|-2=0
∣
z
∣
−
2
=
0
and
∣
z
−
i
∣
−
∣
z
+
5
i
∣
=
0
|z-i|-|z+5i|=0
∣
z
−
i
∣
−
∣
z
+
5
i
∣
=
0
, then
A
x
+
2
y
−
4
=
0
x+2y-4=0
x
+
2
y
−
4
=
0
B
x
2
+
y
−
4
=
0
x^2+y-4=0
x
2
+
y
−
4
=
0
C
x
+
2
y
+
4
=
0
x+2y+4=0
x
+
2
y
+
4
=
0
D
x
2
−
y
+
3
=
0
x^2-y+3=0
x
2
−
y
+
3
=
0
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Solutions · 1
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Official solution
∣
z
−
i
∣
=
∣
z
+
5
i
∣
|z-i|=|z+5i|
∣
z
−
i
∣
=
∣
z
+
5
i
∣
means
z
z
z
lies on the perpendicular bisector of
(
0
,
1
)
(0,1)
(
0
,
1
)
and
(
0
,
−
5
)
(0,-5)
(
0
,
−
5
)
, i.e.
y
=
−
2
y=-2
y
=
−
2
.
∣
z
∣
=
2
⟹
x
2
+
y
2
=
4
⟹
x
2
+
4
=
4
⟹
x
=
0
|z|=2 \implies x^2+y^2=4 \implies x^2+4=4 \implies x=0
∣
z
∣
=
2
⟹
x
2
+
y
2
=
4
⟹
x
2
+
4
=
4
⟹
x
=
0
. So
(
x
,
y
)
=
(
0
,
−
2
)
(x,y)=(0,-2)
(
x
,
y
)
=
(
0
,
−
2
)
, and checking:
x
+
2
y
+
4
=
0
−
4
+
4
=
0
x+2y+4=0-4+4=0
x
+
2
y
+
4
=
0
−
4
+
4
=
0
.
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