If one of the vertices of the square, inscribed in the circle |z−1|=2, is 2+√3i, then other vertices of the square are
A
i√3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(1−√3)+i
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(√3+1)−i
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
−i√3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D−i√3 The given circle is |z−1|=2 where z0=1 is the centre and 2 is radius of the circle. z1 is one of the vertices of the square inscribed in the given circle.
Hence all the other vertices can be obtained by rotating the z1 about z0 by ±π2,π. Thus, z2−z0=(z1−z0)eiπ/2 ⇒z2−1=(2+i√3−1)(i) ⇒z2=i−√3+1 ⇒z2=(1−√3)+i
similarly z4=(√3+1)−i
and z3−z0=(z1−z0)eiπ ⇒z3=z0−(z1−z0)=−i√3