The correct options are
A the maximum value of |2z1+z2| is 4.
B the minimum value of |z1−z2| is 1.
C ∣∣∣z2+1z1∣∣∣≤3
|2z1+z2|≤2|z1|+|z2|=2×1+2=4
∴ Maximum value of |2z1+z2| is 4.
|z1−z2| is least when 0,z1,z2 are collinear.
|z1−z2|≥∣∣|z1|−|z2|∣∣=|1−2|=1
Then min|z1−z2|=1
Again, ∣∣∣z2+1z1∣∣∣≤|z2|+∣∣∣1z1∣∣∣=|z2|+1|z1|=2+11=3⇒∣∣∣z2+1z1∣∣∣≤3
P lies outside |z1|=1
Hence, required minimum distance =OP−r, where O≡(0,0)
=√2−1