AP EAPCET 23-Aug-2021 Shift 1 Paper

Given condition,
(a+bi)3=a−bi
i.e. a3+i3b3+3a2bi−3ab2i2=a−bi
⇒a3−ib3+i3a2b−3ab2=a−bi
(a3−3ab2)−i(b3−3a2b)=a−bi
Equating the coefficient, we obtain
a3−3ab2=a and b3−3a2b=b
⇒a(a2−3b2)=a and b(b2−3a2)=b
⇒a2−3b2=1 .....(1)
and b2−3a2=1.....(2)
Use Eq. (i) in (ii)
b2−3(1+3b2)=1
⇒b2−3−9b2=1‌or‌8‌b2+4=0
This gives b2=−

1

2

∴b=±

i

√2

Now,
a2=1+3b2=1+3−12 =1−

3

2

=−

1

2

∴a=±

i

√2

∴ Possible ordered pairs are
(

i

√2

,

i

√2

), (−

i

√2

,−

i

√2

), (

i

√2

,−

i

√2

), (−

i

√2

,

i

√2

)
Number of ordered pair =4