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Vertices of a Regular Hexagon Circumscribing |z - 1 - i| = root3 | JEE

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Question
If one of the vertices of the regular hexagon circumscribing the circle ∣z−1−i∣=3|z - 1 - i| = \sqrt3 is 2+(1+3)i2 + \left(1+\sqrt3\right)i, then the complex number which is NOT representing any vertex of the hexagon is
A3+i3 + i
B(1+3)i\left(1+\sqrt3\right)i
C−3+i-3 + icorrect
D2+(1−3)i2 + \left(1-\sqrt3\right)i
Solution
Question attachment Step 1: Read off the circle. ∣z−(1+i)∣=3|z - (1+i)| = \sqrt3 has centre C=1+iC = 1 + i and radius 3\sqrt3. Because the hexagon circumscribes this circle, the circle is the hexagon's incircle. Step 2: Find the distance from the centre to a vertex (the circumradius). Check it directly on the given vertex z1=2+(1+3)iz_1 = 2 + \left(1+\sqrt3\right)i:
z1−C=2+(1+3)i−1−i=1+3 i,∣z1−C∣=1+3=2.z_1 - C = 2 + \left(1+\sqrt3\right)i - 1 - i = 1 + \sqrt3\,i, \qquad \left|z_1 - C\right| = \sqrt{1 + 3} = 2 .
 Step 3: Use rotation to get the other vertices. Consecutive vertices of a regular hexagon subtend 2π6=π3\dfrac{2\pi}{6} = \dfrac{\pi}{3} at the centre, so
zk+1−C=(zk−C)e iπ/3,e iπ/3=12+32i.z_{k+1} - C = \left(z_k - C\right)e^{\,i\pi/3}, \qquad e^{\,i\pi/3} = \frac12 + \frac{\sqrt3}{2}i .
 Step 4: Apply it once. With z1−C=1+3iz_1 - C = 1 + \sqrt3 i,
(1+3i)(12+32i)=12+32i+32i+32i2=−1+3 i,\left(1+\sqrt3 i\right)\left(\frac12 + \frac{\sqrt3}{2}i\right) = \frac12 + \frac{\sqrt3}{2}i + \frac{\sqrt3}{2}i + \frac{3}{2}i^{2} = -1 + \sqrt3\,i ,
 so z2=(1+i)+(−1+3i)=(1+3)iz_2 = (1+i) + \left(-1+\sqrt3 i\right) = \left(1+\sqrt3\right)i. That is option (B). Step 5: Keep rotating. Each further multiplication by eiπ/3e^{i\pi/3} gives
z3−C=−2,z4−C=−1−3i,z5−C=1−3i,z6−C=2.z_3 - C = -2, \qquad z_4 - C = -1-\sqrt3 i, \qquad z_5 - C = 1 - \sqrt3 i, \qquad z_6 - C = 2 .
 Step 6: Add C=1+iC = 1+i back to list all six vertices.
z1=2+(1+3)i,z2=(1+3)i,z3=−1+i,z_1 = 2 + \left(1+\sqrt3\right)i, \quad z_2 = \left(1+\sqrt3\right)i, \quad z_3 = -1 + i,
z4=(1−3)i,z5=2+(1−3)i,z6=3+i.z_4 = \left(1-\sqrt3\right)i, \quad z_5 = 2 + \left(1-\sqrt3\right)i, \quad z_6 = 3 + i .
 Step 7: Compare with the options. 3+i3+i is z6z_6, (1+3)i\left(1+\sqrt3\right)i is z2z_2, and 2+(1−3)i2+\left(1-\sqrt3\right)i is z5z_5. The only one missing from the list is −3+I-3 + I. Answer: (3).
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