AIEEE 2003 Math Solved Paper

Given that
|zω|=1
⇒|z||ω|=1

⇒|z|=‌

1

|ω|

and Arg(z)−Arg(ω)=‌

π

2

⇒Arg(‌

z

ω

)=‌

π

2

When argument of a complex number is ‌

π

2

, it means it is making an angle of ‌

π

2

with the real axis in the counterclockwise, so it is along the imaginary axis and positive side of imaginary axis.
So, ‌

z

ω

is a purely imaginary number that means there is no real part in this complex number.
So we can assume,
‌

z

ω

=ki
⇒|‌

z

ω

‌

‌

|=|ki|
⇒|‌

z

ω

‌

‌

|=|k||i|
⇒|‌

z

ω

‌

‌

|=k‌‌ [as |i|=1]
⇒|z|×‌

1

|ω|

=k
⇒|z|×|z|=k‌‌[. as ‌

1

|ω|

=|z|]
⇒|z|2=k
⇒|z|=√k
∴|ω|=‌

1

√k

As ‌

z

ω

is imaginary so we can write,
‌

z

ω

=−‌

z

ω

[When z is imaginary then z=−z ]
⇒zω=−zω
⇒zω=−‌

z

ω

⋅ω⋅ω
⇒zω=−‌

z

ω

⋅|ω|2
⇒zω=−ki⋅(‌

1

√k

)2
⇒zω=−ki⋅‌

1

k

⇒zω=−i