JEE Main 2026 Mathematics question
Let S={z∈ℂ : z²+√6 iz−3=0}. Then is equal to:
Answer: (A)
Solution
z=[−i√6±√(−6+12)]/2=(±√6−i√6)/2=√6(±1−i)/2. Then z²=(6/4)(1∓2i−1)=∓3i, z⁴=−9, z⁸=81. Each of the two roots gives 81, so the sum is 162.
Let S={z∈ℂ : z²+√6 iz−3=0}. Then is equal to:
z=[−i√6±√(−6+12)]/2=(±√6−i√6)/2=√6(±1−i)/2. Then z²=(6/4)(1∓2i−1)=∓3i, z⁴=−9, z⁸=81. Each of the two roots gives 81, so the sum is 162.