JEE Main 2026 — Mathematics Algebra

Given equation is $z^2 + \sqrt{6}iz - 3 = 0$z2+6​iz−3=0.

Rearranging the terms, we get $z^2 - 3 = -\sqrt{6}iz$z2−3=−6​iz.

Squaring both sides, we obtain:

$$(z^2 - 3)^2 = (-\sqrt{6}iz)^2$$(z2−3)2=(−6​iz)2

$$z^4 - 6z^2 + 9 = -6z^2$$z4−6z2+9=−6z2

$$z^4 + 9 = 0$$z4+9=0

$$z^4 = -9$$z4=−9

Squaring again, we get:

$$(z^4)^2 = (-9)^2$$(z4)2=(−9)2

$$z^8 = 81$$z8=81

Since the given equation is a quadratic in $z$z, it has two roots, say $z_1$z1​ and $z_2$z2​.

The discriminant of the quadratic equation is $\Delta = (\sqrt{6}i)^2 - 4(1)(-3) = -6 + 12 = 6 \neq 0$Δ=(6​i)2−4(1)(−3)=−6+12=6=0, which means the roots are distinct.

Thus, the set $S$S contains two distinct elements $z_1$z1​ and $z_2$z2​.

For each root, $z^8 = 81$z8=81.

Therefore, $\displaystyle \sum\limits_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$z∈S∑​z8=z18​+z28​=81+81=162.

Answer: $162$162