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Let complex $z$ such $|z|=1$ Find the maximum of the value $$f(z)=|z^3+3z+2i|$$

I try following $$z=e^{ix}\Longrightarrow z^3=e^{3ix}=\cos{3x}+i\sin{3x}$$ so $$z^3+3z+2i=(\cos{3x}+3\cos{x})+(\sin{3x}+3\sin{x}+2)i$$ so we have $$|z^3+3z+2i|=\sqrt{(\cos{3x}+3\cos{x})^2+(\sin{3x}+3\sin{x}+2)^2}=\sqrt{14+6\cos{3x}\cos{x}+6\sin{3x}\sin{x}+4\sin{3x}+12\sin{x}}=\sqrt{14+6\cos{2x}+4\sin{3x}+12\sin{x}}=\sqrt{20-4\sin^3{x}+24\sin{x}-12\sin^2{x}}$$

even we can use this find maximum, I ask have other more simple method to find this maximum?

math110
  • 93,304

2 Answers2

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Use the complex exponential:

Set $z=\mathrm e^{i\theta}$. Finding the maximum of $\vert f(z)\rvert=\vert z^3+3z+2i\rvert$ amounts to finding the maximum of \begin{align}\vert f(z)\rvert^2&=f(z)\,\overline{\!f(z)\!}\,=(\mathrm e^{3i\theta}+3\mathrm e^{i\theta}+2i)(\mathrm e^{-3i\theta}+3\mathrm e^{-i\theta}-2i)\\ &=14+3(\mathrm e^{2i\theta}+\mathrm e^{-2i\theta})-2i(\mathrm e^{3i\theta}-\mathrm e^{-3i\theta})-6i(\mathrm e^{i\theta}-\mathrm e^{-i\theta})\\&=14+6\cos 2\theta+4\sin3\theta+12\sin\theta \\ &=14+(6-12\sin^2 \theta)+(12\sin\theta-16\sin^3\theta)+12\sin\theta \\ &=4(5+6\sin\theta-3\sin^2 \theta-4\sin^3\theta). \end{align} There remains to find the maximum of the function $p(x)=5+6x-3x^2-4x^3$ on the interval $[-1,1]$. Now $$p'(x)=6(1-x-2x^2)=6(1+x)(1-2x)>0\enspace\text{on }\,(-1,\tfrac12),\; \le 0\enspace\text{otherwise}.$$ Therefor the maximum of $p(x)$ is attained at $x=\frac12$ and $\;p(\frac12)=\frac{27}4$. Therefore $$\max \lvert f(z)\rvert=\frac{3\sqrt 3}{2}.$$

Bernard
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You can find the extrema for the square, for $a,b>0$, $a>b\iff a^2>b^2$

$g(x)=\vert z^3+3z+2i\vert^2=14+6\cos{3x}\cos{x}+6\sin{3x}\sin{x}+4\sin{3x}+12\sin{x}$

$g'(x)=12(-\sin2x+\cos x+\cos3x)=0$

$-\sin2x+\cos x+\cos3x=0$

Etc.

Rafa Budría
  • 7,364