JEE Main 29 June 2022 Shift 2 Solved Paper

Let z=x+iy
∴|z|=√x2+y2
Given, |z|=3
∴√x2+y2=3
⇒x2+y2=9=32
This represent a circle with center at (0,0) and radius =3
Now, given
arg(z−1)−arg(z+1)=‌

π

4

⇒arg(x+iy−1)−arg(x+iy+1)=‌

π

4

⇒arg(x−1+iy)−arg(x+1+iy)=‌

π

4

⇒tan−1(‌

y

x−1

)−tan−1(‌

y

x+1

)=‌

π

4

⇒tan−1(‌

‌ y x−1 − y x+1

1+‌ y x−1 ×‌ y x+1

)=‌

π

4

⇒tan−1(‌

‌ xy+y−xy+y x2−1

x2−1+y2 x2−1

)=‌

π

4

⇒tan−1(‌

xy+y−xy+y

x2−1+y2

)=‌

π

4

⇒‌

2y

x2−1+y2

=tan(‌

π

4

)
⇒2y=x2+y2−1
⇒x2+y2−2y−1=0
⇒x2+(y−1)2=(√2)2
This represent a circle with center at (0,1) and radius √2.
From diagram you can see both the circles do not cut anywhere.