ω=zzˉ−2z+2
∣z−3iz+i∣=1
⇒∣z+i∣=∣z−3i∣
⇒x2+(y+1)2=x2+(y−3)2
⇒y=1
⇒z=x+i,x∈R
ω=(x+i)(x−i)−2(x+i)+2
=x2+1−2x−2i+2
Re(ω)=x2−2x+3=(x−1)2+2
For min (Re(ω)),x=1
⇒ω=2−2i=2(1−i)=22e−i4π
ωn=(22)ne−i4nπ
For real and minimum value of n,
n=4
JEE Main 2021 — Mathematics Algebra
Let z and w be two complex numbers such that w=zzˉ−2z+2,∣z−3iz+i∣=1 and Re(w) has minimum value. Then, the minimum value of n∈N for which wn is real, is equal to _______.
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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