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All Roots of (1+z)^6 + z^6 = 0 Are Collinear | JEE Advanced Complex Numbers

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Question
All roots of the equation (1+z)6+z6=0(1+z)^{6} + z^{6} = 0
Alie on a unit circle with centre at the origin
Blie on a unit circle with centre at (−1,0)(-1, 0)
Clie on the vertices of a regular polygon with centre at the origin
Dare collinearcorrect
Solution
Step 1: Move one term across and take the modulus of both sides.
(1+z)6=−z6  ⟹  ∣1+z∣6=∣−z6∣=∣z∣6.(1+z)^{6} = -z^{6} \implies \left|1+z\right|^{6} = \left|-z^{6}\right| = |z|^{6} .
 Step 2: Both sides are non-negative reals, so we may take the sixth root:
∣z+1∣=∣z∣.|z + 1| = |z| .
 Step 3: Read this geometrically. ∣z+1∣=∣z−(−1)∣|z+1| = |z-(-1)| is the distance of zz from the point (−1,0)(-1,0), and ∣z∣|z| is its distance from the origin (0,0)(0,0). So zz is equidistant from (−1,0)(-1,0) and (0,0)(0,0). Step 4: The set of points equidistant from two fixed points is the perpendicular bisector of the segment joining them - a straight line. Here it is the vertical line
Re⁡(z)=−12.\operatorname{Re}(z) = -\frac12 .
 Step 5: Every root therefore lies on this one line, i.e. the roots are collinear. (They cannot lie on any circle, so (1) and (2) fail, and points on a line are not vertices of a polygon, so (3) fails.) Answer: (4).
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