Question

Xét số phức z= a+bi thoã mãn |z|=1. tính P=2a+4b² khi |z³-z+2| đạt giá trị lớn nhất

Answer 1

Do \(\left|z\right|=1\Rightarrow z=cosx+i.sinx\) với \(\left\{{}\begin{matrix}a=cosx\\b=sinx\end{matrix}\right.\)

\(z^3-z+2=cos3x+i.sin3x-cosx-i.sinx+2\)

\(=cos3x-cosx+2+i\left(sin3x-sinx\right)\)

\(=-2sin2x.sinx+2+i\left(2cos2x.sinx\right)\)

\(=2\left(-sin2x.sinx+1+i.cos2x.sinx\right)\)

\(\Rightarrow A=\left|z^3-z+2\right|=2\sqrt{\left(1-sin2x.sinx\right)^2+cos^22x.sin^2x}\)

\(=2\sqrt{1+sin^22x.sin^2x-2sin2x.sinx+cos^22x.sin^2x}\)

\(=2\sqrt{1+sin^2x-4sin^2x.cosx}=2\sqrt{2-cos^2x-4cosx\left(1-cos^2x\right)}\)

\(=2\sqrt{2-cos^2x-4cosx+4cos^3x}=2\sqrt{4a^3-a^2-4a+2}\)

\(A_{min}\) khi \(f\left(a\right)=4a^3-a^2-4a+2\) đạt min

\(f'\left(a\right)=12a^2-2a-4=0\Rightarrow\left[{}\begin{matrix}a=-\frac{1}{2}\\a=\frac{2}{3}\end{matrix}\right.\)

Dựa vào BBT, ta thấy \(f\left(a\right)\) min khi \(a=\frac{2}{3}\) \(\Rightarrow b^2=1-a^2=\frac{5}{9}\)

\(\Rightarrow P=\frac{32}{9}\)