JEE Main 16 March 2021 Shift 1 Solved Paper

Given, w=zz−2z+2
|‌

z+i

z−3i

|=1
⇒‌‌|z+i|=|z−3i|
Let z=x+iy
⇒‌‌|x+i(y+1)|=|x+i(y−3)|
⇒‌‌x2+(y+1)2=x2+(y−3)2
⇒‌‌2y+1=−6y+9
∴‌‌y=1
Now, w=zz−2z+2
w=|z|2−2z+2
⇒‌w=x2+y2−2(x+iy)+2
⇒‌w=(x2+y2−2x+2)+i(−2y)
⇒‌w=(x2+1−2x+2)+i(−2)
w=(x−1)2+2−2i
Re(w) has minimum value.
So, (x−1)2+2 is minimum when x=1
∴‌‌W=2−2i
=2(1−i)
=2√2(‌

1

√2

−‌

i

√2

)
w=2√2e−iπ∕4
Now, ‌‌wn=(2√2)ne‌

−inπ

4

=(2√2)n[cos(‌

nπ

4

)−i‌i‌sin(‌

nπ

4

)]
This has to be zero for wn to be real.
So, sin(‌

nπ

4

)=0
⇒‌‌‌

nπ

4

=0,π,2π,3π...
⇒‌‌n=0,4,8,12...
The minimum value of n is 4(n∈N).