JEE Advanced 2020 · Paper 1, Mathematics question 9

Let $w=z+\frac12$. Then $z^2+z+1=w^2+\frac34$ and the condition is $\left|w^2+\frac34\right|=1$.

(C) By the triangle inequality, $|w|^2\ge1-\frac34=\frac14$, so $|w|\ge\frac12$. True.

(A) Take $w^2+\frac34=-1$, i.e. $w=\frac{\sqrt7}2i$ and $z=-\frac12+\frac{\sqrt7}2i$. Then $|z^2+z+1|=1$, so $z\in S$, but $|w|=\frac{\sqrt7}2>\frac12$. False.

(B) Similarly $|w|^2\le1+\frac34=\frac74$, so $|z|\le\frac12+\frac{\sqrt7}2\approx1.82\le2$. True.

(D) $w^2$ can be any point of the circle $|u+\frac34|=1$, which gives infinitely many $z$. False.

(B), (C)