Câu hỏi:
-
A.
\(z – \overline z = 2a\). -
B.
\(z + \overline z = 2bi\). -
C.
\(|{z^2}| = |z{|^2}\). -
D.
\(z.\overline z = {a^2} + {b^2}\).
Lời giải tham khảo:
Đáp án đúng: D
Đặt z = a + bi \(a,b \in \mathbb{Z}\)
\(\begin{array}{l}z – \overline z = a + bi – \left( {a – bi} \right) = 2bi\\z + \overline z = a + bi + \left( {a – bi} \right) = 2a\\\left| {{z^2}} \right| = \left| {{{\left( {a + bi} \right)}^2}} \right| = \left| {{a^2} – {b^2} + 2abi} \right|\\\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {{a^2} – {b^2}} \right)}^2} + 4{a^2}{b^2}} \\\,\,\,\,\,\,\,\,\, = \sqrt {{a^4} + 2{a^2}{b^2} + {b^4}} \\\,\,\,\,\,\,\,\,\, = \sqrt {{{\left( {{a^2} + {b^2}} \right)}^2}} = {a^2} + {b^2}\\z\overline z = (a + bi)\left( {a – bi} \right)\\\,\,\,\,\,\,\, = {a^2} – {b^2}{i^2} = {a^2} + {b^2}\end{array}\)
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