How to find maximum value of $| Z| $ if: $$ \Big| Z-\dfrac{4}{Z} \Big|=2; $$
Where $Z$ is a complex mumber
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Did you mean the maximum of $,|z|,$ ? – DonAntonio Mar 28 '13 at 12:56
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I have found another answer here. http://goo.gl/v7L9Q – ABC Mar 28 '13 at 13:33
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Answer given is $|Z|<1+\sqrt5$ – ABC Mar 28 '13 at 14:42
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From here, $$\left|z-\frac4z\right|\ge |z|-\left|\frac4z\right|$$
So, $|z|-\frac4{|z|}\le 2$
$\implies |z|^2-2|z|-4\le 0\implies (|z|-1)^2\le 5$
$\implies -\sqrt 5\le |z|-1\le \sqrt5\implies 1-\sqrt5\le |z|\le 1+\sqrt5$ $\implies 0< |z|\le 1+\sqrt5$
lab bhattacharjee
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how do you show that the value $|z|=1+\sqrt 5$ is actually attained for some $z$? – suissidle Mar 28 '13 at 17:59
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1@suissidle,if $|z|=1+\sqrt 5, |z|-\frac4{|z|}=1+\sqrt5-(\sqrt5-1)=2$ – lab bhattacharjee Mar 28 '13 at 18:05
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Write $Z=re^{i\phi}$, then
$$ 4 = |Z-\frac{4}{Z}|^2 \\= (re^{i\phi}-\frac{4}{r}e^{-i\phi})(re^{-i\phi}-\frac{4}{r}e^{i\phi}) \\= r^2-8cos(2\phi)+\frac{16}{r^2} $$
Write $cos(2\phi)=c$ (this parameter varies freely in $[-1,1]$ given an appropriate choice of $\phi$) and maximize $r$ subject to the constraint $$ r^2-8c+\frac{16}{r^2}=4 $$ now, write $s=r^2$: $$ s^2-2(4c+2)s+16=0 \\ s\in{4c+2\pm\sqrt{(4c+2)^2-16}} $$ $s$ is thus maximized for $c=1$, and $$ |Z|_{max} = r_{max}\\ = \sqrt{s_{c=1}} \\ = \sqrt{6+2\sqrt{5}} \\ = \sqrt{1+2\sqrt{5}+5} \\ = 1+\sqrt5 $$
suissidle
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Won't the last two terms in brackets get multiplied to $\frac{16}{r^2}$??. – ABC Mar 28 '13 at 12:26
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right, edited accordingly (i first did the calculation with $\frac{2}{Z}$ and made a mistake while updating the result) – suissidle Mar 28 '13 at 12:28
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Setting $z=Z/2$ we have $$ |Z-4/Z|=2 \iff |z-z^{-1}|=1. $$ So the problem is to find $$ R=2\max_{z \in M}|z| \ \text{ where }\ M:=\{z \in \mathbb{C}:\ |z-z^{-1}|=1\}. $$ We have $$ M=\{re^{i\theta}:\ r>0, \ -\pi \le \theta \le\pi,\ r^4-(1+2\cos\theta)r^2+1=0\}. $$ Solving $$ r^4-(1+2\cos\theta)r^2+1=0,\ r>0,\ -\pi \le \theta \le\pi, $$ we obtain \begin{eqnarray} r_1^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta+\sqrt{(1+2\cos\theta)^2-1}\right),\ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2},\\ r_2^2(\theta)&=&\frac{1}{2}\left(1+2\cos\theta-\sqrt{(1+2\cos\theta)^2-1}\right),\ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}. \end{eqnarray} Since $$ r_1^2(\theta)\ge r_2^2(\theta) \quad \forall \theta \in [-\pi/2,\pi/2], $$ it follows that $$ R=2\max_{-\pi/2\le \theta \le \pi/2}r_1(\theta)=\max_{1 \le t \le 3}\sqrt{2[t+\sqrt{t^2-1}]}=\sqrt{2(3+2\sqrt{2})}. $$
HorizonsMaths
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