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