Question

# Let S be the set of all complex numbers 𝑧 satisfying $$|z^2+z+1|=1$$ Then which of the following statements is/are TRUE?

A
z+1212 for all zϵS
B
|z|2 for all zϵS
C
z+1212 for all zϵS
D
The set S has exactly four elements

Solution

## The correct options are C $$|z|\le 2$$ for all $$z\,\epsilon\, S$$ D $$\left|z+\dfrac{1}{2}\right|\ge \dfrac{1}{2}$$ for all $$z\,\epsilon \,S$$$$| z^2 + z +1| = 1 = | \left| \left( z + \dfrac{1}{2} \right)^2 + \dfrac{3}{4}\right|$$$$1 \le \left| z + \dfrac{1}{2}\right|^2 + \dfrac{3}{4}$$$$\left| z + \dfrac{1}{2} \right|^2 \ge \dfrac{1}{4}$$$$\left| z + \dfrac{1}{2} \right| \le -\dfrac{1}{2}$$ or $$\left| z + \dfrac{1}{2} \right| \ge \dfrac{1}{2}$$        ....(C)Now$$\left|\left(z+\dfrac{1}{2}\right)^2 + \dfrac{3}{4}\right| = |z^2+z+1| = 1$$$$\left|\left|z + \dfrac{1}{2}\right|^2 - \dfrac{3}{4}\right| \le 1$$$$-1 \le \left|z+\dfrac{1}{2} \right|^2 -\dfrac {3}{4} \le 1$$$$\dfrac {-1}{4} \le \left|z+\dfrac{1}{2} \right|^2 \le \dfrac{7}{4}$$$$\left | z+\dfrac{1}{2}\right| \le \dfrac{\sqrt{7}}{2}$$$$|z| \le \dfrac{\sqrt{7}}{2} - 1$$$$|z| \le \dfrac{\sqrt{7} - 1}{2} \le \dfrac{4}{2}$$$$\boxed{|z| \le 2}$$       ...(B)Mathematics

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