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Câu hỏi: Cho $z+\dfrac{1}{z}=-1$. Tính $P=\left| {{z}^{2023}}+\dfrac{1}{{{z}^{2023}}} \right|$
A. $P=\sqrt{2}$.
B. $P=1$.
C. $P=0$.
D. $P=-1$.
A. $P=\sqrt{2}$.
B. $P=1$.
C. $P=0$.
D. $P=-1$.
$z+\dfrac{1}{z}=-1\Leftrightarrow {{z}^{2}}+z+1=0\Leftrightarrow \left( z-1 \right)\left( {{z}^{2}}+z+1 \right)=0\Leftrightarrow {{z}^{3}}-1=0\Leftrightarrow {{z}^{3}}=1$.
Khi đó $P=\left| {{z}^{2023}}+\dfrac{1}{{{z}^{2023}}} \right|=\left| {{\left( {{z}^{3}} \right)}^{674}}z+\dfrac{1}{{{\left( {{z}^{3}} \right)}^{674}}z} \right|=\left| z+\dfrac{1}{z} \right|=1$.
Khi đó $P=\left| {{z}^{2023}}+\dfrac{1}{{{z}^{2023}}} \right|=\left| {{\left( {{z}^{3}} \right)}^{674}}z+\dfrac{1}{{{\left( {{z}^{3}} \right)}^{674}}z} \right|=\left| z+\dfrac{1}{z} \right|=1$.
Đáp án B.