Cho số phức \(w=x+yi\), \(\left( x\,,\,y\in \mathbb{R} \right)\) thỏa mãn điều kiện \(\left| {{w}^{2}}+4 \right|=2\left| w \right|\). Đặt \(P=8\left( {{x}^{2}}-{{y}^{2}} \right)+12...

* Hướng dẫn giải

Ta có \({{w}^{2}}+4\)\(={{\left( x+yi \right)}^{2}}+4\)\(={{x}^{2}}-{{y}^{2}}+2xyi+4\)\(\Rightarrow \left| {{w}^{2}}+4 \right|=\sqrt[{}]{{{\left( {{x}^{2}}-{{y}^{2}}+4 \right)}^{2}}+4{{x}^{2}}{{y}^{2}}}\)

Do đó \(\left| {{w}^{2}}+4 \right|=2\left| w \right|\)\(\Leftrightarrow \sqrt[{}]{{{\left( {{x}^{2}}-{{y}^{2}}+4 \right)}^{2}}+4{{x}^{2}}{{y}^{2}}}=2\sqrt[{}]{{{x}^{2}}+{{y}^{2}}}\)\(\Leftrightarrow {{\left( {{x}^{2}}-{{y}^{2}}+4 \right)}^{2}}+4{{x}^{2}}{{y}^{2}}=4\left( {{x}^{2}}+{{y}^{2}} \right)\)\(\Leftrightarrow {{x}^{4}}+{{y}^{4}}-2{{x}^{2}}{{y}^{2}}+8\left( {{x}^{2}}-{{y}^{2}} \right)+16+4{{x}^{2}}{{y}^{2}}=4\left( {{x}^{2}}+{{y}^{2}} \right)\)

\(\Leftrightarrow {{x}^{4}}+{{y}^{4}}+2{{x}^{2}}{{y}^{2}}-4\left( {{x}^{2}}+{{y}^{2}} \right)+4+8\left( {{x}^{2}}-{{y}^{2}} \right)+12=0\)\(\Leftrightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-4\left( {{x}^{2}}+{{y}^{2}} \right)+4+8\left( {{x}^{2}}-{{y}^{2}} \right)+12=0\)\(\Leftrightarrow {{\left( {{x}^{2}}+{{y}^{2}}-2 \right)}^{2}}+8\left( {{x}^{2}}-{{y}^{2}} \right)+12=0\)

\(\Leftrightarrow 8\left( {{x}^{2}}-{{y}^{2}} \right)+12=-{{\left( {{x}^{2}}+{{y}^{2}}-2 \right)}^{2}}\)\(\Leftrightarrow P=-{{\left( {{\left| w \right|}^{2}}-2 \right)}^{2}}\).