Google
Câu hỏi: Cho số phức $z=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$. Tìm số phức $w=1+z+{{z}^{2}}$.
A. $2-\sqrt{3}i$.
B. $0$.
C. $1$.
D. $-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
A. $2-\sqrt{3}i$.
B. $0$.
C. $1$.
D. $-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$.
Ta có ${{z}^{2}}={{\left( -\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)}^{2}}=\dfrac{1}{4}-\dfrac{3}{4}-\dfrac{\sqrt{3}}{2}i=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i$.
Vậy $w=1+z+{{z}^{2}}=1-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i=0$.
Vậy $w=1+z+{{z}^{2}}=1-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i=0$.
Đáp án B.