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India LIMIT
2019 LIMIT
2019 LIMIT Category A
Problem 11
condition for z/(1-z^2)
condition for z/(1-z^2)
Source: LIMIT 2019 CAS1 P11
April 28, 2021
complex numbers
algebra
Problem Statement
z
z
z
is a complex number and
∣
z
∣
=
1
|z|=1
∣
z
∣
=
1
and
z
2
≠
1
z^2\ne1
z
2
=
1
. Then
z
1
−
z
2
\frac z{1-z^2}
1
−
z
2
z
lies on
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
a line not through origin
<span class='latex-bold'>(A)</span>~\text{a line not through origin}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
a line not through origin
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
|z|=2
<span class='latex-bold'>(B)</span>~\text{|z|=2}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
|z|=2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
x
−
axis
<span class='latex-bold'>(C)</span>~x-\text{axis}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
x
−
axis
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
y
−
axis
<span class='latex-bold'>(D)</span>~y-\text{axis}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
y
−
axis
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