JEE Main 2026 — Complex Number Question with Solution

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

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

Squaring both sides, we obtain:

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

$z^4-6z^2+9=-6z^2$

$z^4+9=0$

$z^4=-9$

Squaring again, we get:

$(z^4{)}^2=(-9{)}^2$

$z^8=81$

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

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

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

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

Therefore, $\sum _{z\in S}{z}^8=z_1^8+z_2^8=81+81=162$.

Answer: $162$