Let S be the set of all complex numbers z sat | vedclass

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(B) Given $|z^2+z+1|=1$.We can write this as $|(z+\frac{1}{2})^2 + \frac{3}{4}| = 1$.Using the triangle inequality $|a+b| \leq |a| + |b|$,we have $1 =

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$B, C$

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(B) Given $|z^2+z+1|=1$.We can write this as $|(z+\frac{1}{2})^2 + \frac{3}{4}| = 1$.Using the triangle inequality $|a+b| \leq |a| + |b|$,we have $1 = |(z+\frac{1}{2})^2 + \frac{3}{4}| \leq |z+\frac{1}{2}|^2 + \frac{3}{4}$,which implies $|z+\frac{1}{2}|^2 \geq \frac{1}{4}$,so $|z+\frac{1}{2}| \geq \frac{1}{2}$. Thus,$(C)$ is true.Also,$|z^2+z| = |(z^2+z+1)-1| \leq |z^2+z+1| + |-1| = 1+1 = 2$.Since $|z^2+z| = |z||z+1| \leq 2$,for large $|z|$,$|z|^2 \approx |z^2+z| \leq 2$,so $|z| \leq 2$. Thus,$(B)$ is true.Since the equation $|z^2+z+1|=1$ represents a curve in the complex plane,the set $S$ is infinite,so $(D)$ is false.Therefore,the correct statements are $(B)$ and $(C)$.

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