Question

Cho số phức \(z=a+bi\) thỏa \(\left|z\right|=1\) . Tính \(P=2a+4b^2\) Khi \(\left|z^3-z+2\right|\) đạt giá trị lớn nhất.

A. \(P=4\)

B. \(P=2-\sqrt{2}\)

C. \(P=2\)

D. \(P=2+\sqrt{2}\)

Answer 1

\(\left|z\right|=1\Rightarrow z=cosx+i.sinx\)

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

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

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

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

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

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

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

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

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

\(A_{max}\) khi \(4cos^3x-cos^2x-4cosx+2\) đạt max

Xét hàm \(f\left(t\right)=4t^3-t^2-4t+2\) trên \(\left[-1;1\right]\)

\(f'\left(t\right)=12t^2-2t-4=0\Rightarrow\left[{}\begin{matrix}t=-\frac{1}{2}\\t=\frac{2}{3}\end{matrix}\right.\)

\(\Rightarrow f\left(t\right)\) đạt max tại \(t=-\frac{1}{2}\) hay \(A_{max}\) khi \(a=cosx=-\frac{1}{2}\)

\(\Rightarrow b^2=sin^2x=1-cos^2x=\frac{3}{4}\)

\(\Rightarrow P=2a+4b^2=-1+3=2\)