$C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$

Question

My try:
I applied $|z^3-z+2|\leq|z|^3+|-z|+|2|$,i got $f(z)\leq 4$ but book says my answer is wrong.

Correct answer is $|f(z)|$ is maximum when $z=w,$where $w$ is cube root of unity and $|f(z)|=\sqrt{13}$.

I do not know how to get correct answer.

Do you mean $\mathbb{C}$ instead of $C$ for the domain of $f$? As in the latter case, your function is only defined for on complex number. — Hetebrij, Mar 13 '16 at 16:30
try to use $z=e^{i\theta}$, and then try to find max value of function in terms of $\theta$ — Kerr, Mar 13 '16 at 17:03

score 3 · Accepted Answer · answered Mar 17 '16 at 06:03

3

Let $\cos\theta=c,\sin\theta=s$ and $z=c+si$.

Then, since $$\begin{align}z^3-z+2&=(c+si)^3-(c+si)+2\\&=c^3+3c^2si-3cs^2-s^3i-c-si+2\\&=(c^3-3cs^2-c+2)+(3c^2s-s^3-s)i\end{align}$$ we have $$\begin{align}f(z)&=\sqrt{(c^3-3cs^2-c+2)^2+(3c^2s-s^3-s)^2}\\&=\sqrt{c^6+3 c^4 s^2-2 c^4+4 c^3+3 c^2 s^4+c^2-12 c s^2-4 c+s^6+2 s^4+s^2+4 }\\&=\sqrt{(c^6+3c^4s^2+3c^2s^4+s^6)+(s^2+c^2)-2(c^4-s^4)-12cs^2+4c^3-4c+4}\\&=\sqrt{(c^2+s^2)^3+1-2(c^2+s^2)(c^2-s^2)-12c(1-c^2)+4c^3-4c+4}\\&=\sqrt{1+1-2\cdot 1\cdot (c^2-(1-c^2))-12c(1-c^2)+4c^3-4c+4}\\&=\sqrt{16c^3-4c^2-16c+8}\end{align}$$

Here, let $g(t)=16t^3-4t^2-16t+8$. Then, $$g'(t)=0\iff t=-\frac 12,\frac 23.$$ Hence, the maximum value of $g(t)$ for $-1\le t\le 1$ is $$\max\{g(-1/2),g(1)\}=\max\{13,4\}=13.$$

Therefore, the maximum value of $f(z)$ is $f((-1\pm\sqrt 3\ i)/2)=\sqrt{13}$.

answered Mar 17 '16 at 06:03

mathlove

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To mathlove how can we calculate $\displaystyle x = -\frac{1}{2}\pm \frac{\sqrt{3}}{2}i.,$ Thanks – juantheron Mar 17 '16 at 06:43
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@juantheron: We have $g(-1/2)=13$. So, the max is attained when $\cos\theta=c=-1/2$. – mathlove Mar 17 '16 at 06:47
Thanks mathlove, would you like to explain me how can we get Imaginary parts. – juantheron Mar 17 '16 at 06:51
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@juantheron: Using $\cos^2\theta+\sin^2\theta=1$, the imaginary part is $\sin\theta=\pm\sqrt{1-(-1/2)^2}=\pm\sqrt 3/2$. – mathlove Mar 17 '16 at 06:52

$C$ is a complex number.$f:C\to R$ is defined by $f(z)=|z^3-z+2|.$Find the maximum value of $f(z)$ if $|z|=1.$

1 Answers1