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

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Question

$\left|z^2+z+1\right|=1$

$\Rightarrow  \left|\left(z+\frac{1}{2}\right)^2+\frac{3}{4}\right|=1$

$\Rightarrow  \left|\left(z+\frac{1}{2

Vedclass · Accepted Answer

$B,C$

Vedclass · Answer

$\left|z^2+z+1ight|=1$

$\Rightarrow  \left|\left(z+\frac{1}{2}ight)^2+\frac{3}{4}ight|=1$

$\Rightarrow  \left|\left(z+\frac{1}{2}ight)^2+\frac{3}{4}ight| \leq\left|z+\frac{1}{2}ight|^2+\frac{3}{4}$

$\Rightarrow  1 \leq\left|z+\frac{1}{2}ight|^2+\frac{3}{4} \Rightarrow\left|\left(z+\frac{1}{2}ight)ight|^2 \geq \frac{1}{4}$

$\Rightarrow  \left|z+\frac{1}{2}ight| \geq \frac{1}{2}$

$	ext { also }  \left|\left(z^2+zight)+1ight|=1 \geq|| z^2+z|-1|$

$\Rightarrow  \left|z^2+zight|-1 \leq 1$

$\Rightarrow  \left|z^2+zight| \leq 2$

$\Rightarrow  \left\|z^2|-| zight\| \leq\left|z^2+zight| \leq 2$

$\Rightarrow  \left|r^2-right| \leq 2$

$\Rightarrow  \quad r =|z| \leq 2 ; \forall z \in S$

Also we can always find root of the equation $z^2+z+1=e^{i 	heta} ; \forall 	heta \in R$

Hence set ' $S$ ' is infinite

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