257. Seeking roots of 2th equation with complex coef.

by Virgil Nicula, Apr 1, 2011, 9:18 PM

Preliminary. Let $az^2+bz+c=0$ be a second degree polynomial equation with complex coefficients, i.e. $\{a,b,c\}\subset\mathbb C$ and $a\ne 0$ . Suppose w.l.o.g. that exists at least a nonreal number

in the set $\{a,b,c\}$ and $\Delta=b^2-4ac=u+iv$ , where $\{u,v\}\subset\mathbb R$ and $v\ne 0$ . Observe that $az^2+bz+c=0\iff (2az+b)^2=\Delta\ (*)$ . Therefore, at first solve

the equation $y^2=u+iv$ . Prove easily that $y_{1,2}=\pm w$ , where $w=\sqrt{\frac {r+u}{2}}+i\cdot\mathrm{sgn}(v)\cdot\sqrt {\frac {r-v}{2}}$ and $r=|\Delta|$ . At second from the relation $(*)$ obtain that $z_{1,2}=\frac {-b\pm w}{2a}$ .

Example. Solve the equation $z^2-8(1-i)\cdot z+(63-16\cdot i)=0$ .

$\blacktriangleright\ \Delta^{\prime}=16(1- i)^2-(63-16\cdot i)=-63-16\cdot i\ ;$

$\blacktriangleright\ y^2=-63-16\cdot i\implies\ \left\{\begin{array}{c} r=\sqrt {63^2+16^2}=65\\\\ y_{1,2}=\pm\left(\sqrt {\frac {65-63}{2}}-i\cdot\sqrt {\frac {65+63}{2}}\right)=\pm(1-8\cdot i)\end{array}\right\|\ ;$

$\blacktriangleright\ z_{1,2}=4(1- i)\pm (1-8\cdot i)\ \implies\ z_1=5-12\cdot i$ and $z_2=3+4\cdot i$ .

Remark. With the Viete's relations verify that $S=z_1+z_2=8(1-i)=-\frac ba$ and $P=z_1z_2=63-16\cdot i=\frac ca$ .

Proposed problem. Let $z_{1,2}\in\mathbb C$ be the roots of the equation with complex coefficients $az^2+bz+c=0\ ,\ a\ne 0$ . Prove that $\boxed{\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}}$ .
Proof. $\left\{\begin{array}{cccc} \left|z_1-z_2\right|^2=\left|\left(z_1+z_2\right)^2-4z_1z_2\right| & \implies & \left|z_1-z_2\right|^2=\left|\frac {b^2-4ac}{a^2}\right| & (1)\\\\ \boxed{\left|z_1+z_2\right|^2+\left|z_1-z_2\right|^2=2\left(\left|z_1\right|^2+\left|z_2\right|^2\right)} & \stackrel{(1)}{\implies} & \left|z_1\right|^2+\left|z_2\right|^2=\frac 12\cdot \left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{|a|^2}\right) & (2)\\\\ \left(\left|z_1\right|+\left|z_2\right|\right)^2=\left|z_1\right|^2+\left|z_2\right|^2+2\left|z_1z_2\right| & \stackrel{(2)}{\implies} & \left(\left|z_1\right|+\left|z_2\right|\right)^2=\frac 12\cdot\left(\left|\frac ba\right|^2+\frac {\left|b^2-4ac\right|}{2|a|^2}\right)+2\cdot\left|\frac ca\right| & (3)\end{array}\right\|$ $\stackrel{(3)}{\implies}$ $\left|z_1\right|+\left|z_2\right|=\sqrt {\frac {|b|^2+|b^2-4ac|+4|ac|}{2|a|^2}}$ .

Example. Let $\left\{z_1,z_2\right\}\subset\mathbb C$ be the roots of the equation $z^2-2iz+m=0$ , where $m\in\mathbb C$ is a parameter. Ascertain the values of $m$ for which $\left|z_1\right|+\left|z_2\right|=2$ .

Proof. For $a:=1\ ,\ b:=-2i$ and $c:=m\in\mathbb C$ obtain that $\left|z_1\right|+\left|z_2\right|=2\iff$ $\sqrt {\frac {4+4|1+m|+4|m|}{2}}=2\iff$ $|m|+|m+1|=1$ . Denote the

points $M(m)\ ,\ A(0)\ ,\ B(-1)$ . Therefore, $|m|+|m+1|=1\iff$ $MA+MB=AB\iff$ $M\in[AB]\iff m\in\mathbb R\ ,\ m\in [-1,0]$ . Nice exercise !
This post has been edited 29 times. Last edited by Virgil Nicula, Nov 22, 2015, 9:39 AM

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    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
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404398
  • Total comments: 37
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