Bài 4.12 trang 202 SBT giải tích 12

Đề bài

Cho \(z = a + bi\). Chứng minh rằng:

a) \({z^2} + {\left( {\overline z } \right)^2} = 2({a^2} – {b^2})\)

b) \({z^2} – {\left( {\overline z } \right)^2} = 4abi\)

c) \({z^2}{\left( {\overline z } \right)^2} = {\left( {{a^2} + {b^2}} \right)^2}\)

Lời giải chi tiết

Ta có: \({z^2} = {(a + bi)^2} = {a^2} – {b^2} + 2abi\)

\({(\overline  z)^2} = {(a – bi)^2} = {a^2} – {b^2} – 2abi\)

a) \({z^2} + {\left( {\overline z } \right)^2}\) \( = {a^2} – {b^2} + 2abi + {a^2} – {b^2} – 2abi\) \( = 2\left( {{a^2} – {b^2}} \right)\).

b) \({z^2} – {\left( {\overline z } \right)^2}\)\( = {a^2} – {b^2} + 2abi – {a^2} + {b^2} + 2abi\)\( = 4abi\).

c) \({z^2}.{\left( {\overline z } \right)^2} = {\left( {z.\overline z } \right)^2}\) \( = {\left[ {\left( {a + bi} \right)\left( {a – bi} \right)} \right]^2} \) \(= {\left( {{a^2} – {b^2}{i^2}} \right)^2} = {\left( {{a^2} + {b^2}} \right)^2}\).

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