256. An extremum problems with complex numbers.

by Virgil Nicula, Apr 1, 2011, 1:37 PM

Proposed problem. Let $z\in\mathbb C^*$ be a complex number such that $\left|2z +\frac{1}{z}\right|=1$ and $\arg (z) =\theta$ . Ascertain the minimum value of $8\cdot \sin^2\theta$ .

Proof. Denote $z=\rho\cdot (\cos\theta +i\cdot \sin\theta )$ , where $\rho >0$ and $\theta\in \left[0,2\pi\right)$ . Thus, $z\cdot\overline z=\rho^2$ , $z+\overline z=2\rho\cos\theta$ and $z^2+\overline z^2=(z+\overline z )^2-2z\cdot\overline z=$

$4\rho^2\left(1-\sin^2\theta \right)-2\rho^2\implies$ $z^2+\overline z^2=2\rho^2-4\rho^2\sin^2\theta$ . Hence $\left|2z +\frac{1}{z}\right|=1\iff$ $\left|2z^2+1\right|=|z|\iff$ $\left(2z^2+1\right)\left(2\overline z^2+1\right)=z\overline z\iff$

$4\left(z\overline z\right)^2+2\left(z^2+\overline z^2\right)+1=z\overline z\iff$ $4\rho^4+2\left(2\rho^2-4\rho^2\sin^2\theta\right)+1=\rho^2\iff$ $8\sin^2\theta =\frac {4\rho^4+3\rho^2+1}{\rho^2}=$ $4\rho^2+\frac {1}{\rho^2}+3\ge 4+3=7$ .

Remark. $\left|2z+\frac 1z\right|=1\iff\left(2z+\frac{1}{z}\right)\left(2\overline{z}+\frac{1}{\overline{z}}\right)=1\iff$ $4|z|^{2}+\frac{1}{|z|^{2}}+2\left(\frac{z}{\overline{z}}+\frac{\overline{z}}{z}\right)=1\iff$

$4|z|^{2}+\frac{1}{|z|^{2}}+4\cos 2\theta =1$ $\Longrightarrow$ $1-4\cos 2\theta\geq 4$ by AM-GM inequality $\iff$ $4\left(1-2\sin^2\theta\right)\le -3\iff$ $8\sin^{2}\theta\geq 7$ .


An easy extension. Let $z\in\mathbb C^*$ be a complex number such that $\left|az +\frac{b}{z}\right|=1$ and $\arg (z) =\theta$ ,

where $\{a,b\}\subset\mathbb R^*$ and $ab>\frac 14$ . Ascertain the maximum of $|z|$ and prove that $\cos^2\theta\le\frac {1}{4ab}$ .

Remark. Suppose $\{a,b,c\}\subset\mathbb R$ , where $a>0$ and $b>0$ . Observe that if denote $|z|=\rho$ , then $\left|az^2+b\right|=|z+c|\iff$ $\left(az^2+b\right)\left(a\overline z^2+b\right)=$

$(z+c)\left(\overline z+c\right)\iff$ $a^2\rho^4+ab\left[\left(z+\overline z\right)^2-2\rho^2\right]+b^2=$ $\rho^2+c\left(z+\overline z\right)+c^2\iff$ $a^2\rho^4-\left(2ab+1\right)\rho^2+ab\left(z+\overline z\right)^2-c\left(z+\overline z\right)=$

$c^2-b^2\iff$ $\left(a\rho^2-\frac {2ab+1}{2a}\right)^2+\left[\sqrt {ab}\cdot\left(z+\overline z\right)-\frac {c}{2\sqrt {ab}}\right]^2=$ $c^2-b^2+\left(\frac {2ab+1}{2a}\right)^2+\frac {c^2}{4ab}\implies$ $\left\{\begin{array}{c} \left|a\rho^2-\frac {2ab+1}{2a}\right|\le\sqrt{c^2-b^2+\left(\frac {2ab+1}{2a}\right)^2+\frac {c^2}{4ab}}\\\\ \left|\sqrt {ab}\cdot\left(z+\overline z\right)-\frac {c}{2\sqrt {ab}}\right|\le \sqrt{c^2-b^2+\left(\frac {2ab+1}{2a}\right)^2+\frac {c^2}{4ab}}\end{array}\right\|$ .

Some examples. Let $z\in\mathbb C^*$ be a complex number with $\rho=|z|$ and $\mathrm{Re}(z)=\frac 12\cdot\left(z+\overline z\right)$ . Ascertain the maximum for $\rho$ and for $\mathrm{Re}(z)$ , where :

$1\blacktriangleright\ \left|z^2+1\right|=a|z|\ ;\ \ 2\blacktriangleright\ \left|z^2+1\right|=$ $|z-1|\ ;\ \ 3\blacktriangleright\ \left|z^2+1\right|=|4z+3|\ ;\ \ 4\blacktriangleright\ \left|z^2+1\right|=2\cdot |z+1|\ .$
This post has been edited 31 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:41 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404396
  • Total comments: 37
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