Let z=x+iy. From 4z2+zˉ=0, separating real and imaginary parts:
Real: 4x2−4y2+x=0, Imaginary: y(8x−1)=0.
Case 1: y=0
⇒x(4x+1)=0
⇒z=0 or z=−41.
Case 2: x=81
⇒161−4y2+81=0
⇒y2=643.
So ∣z∣2=641+643=161.
∑∣z∣2=0+161+161+161=163
Let S={z∈C:4z2+zˉ=0}. Then z∈S∑∣z∣2 is equal to:
Held on 22 Jan 2026 · Verified 6 Jul 2026.
645
161
163
647
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