complex number (1 Viewer)

juantheron · Dec 20, 2016

$If $a,b,c$ are three distinct non zero complex numbers such that$

$|a| = |b| = |c|$ and the equation $az^2+bz+c=0$ has a roots whose$$

$modulus is $1\;,$ Then relation between $a,b,c$ is$

calamebe · Dec 20, 2016

Well let the roots be cisx and cisy, because they're easy to write. So we know that cisx+cisy = -b/a and cisx*cisy=cis(x+y)=c/a.
Squaring the first equation, b^2/a^2 = cis(x)^2 +2cis(x)*cis(y) + cis(y)^2 = cis(2x) + 2cis(x+y) + cis(2y). So as cis(x+y) = c/a:
(b^2/a^2) - 2c/a = cis(2x) + cis(2y).

And I'm stuck so I'll just leave this here in case it helps anyone

juantheron · Dec 23, 2016

$Let $z_{1},z_{2}$ be two roots of $az^2+bz+c=0.$ So $z_{1}+z_{2} = -\frac{b}{a}\;,z_{1}\cdot z_{2} = \frac{c}{a}$

$Given $|z_{1}| = |z_{2}| = 1$ and $|a| = |b| = |c| = 1$

$So $|z_{1}+z_{2}|^2 = 1\Rightarrow (z_{1}+z_{2})\cdot (\bar{z_{1}}+\bar{z_{2}}) = 1\Rightarrow z_{1}\bar{z_{2}}+\bar{z_{1}}z_{2} = -1$

$Now since $z_{1},z_{2}$ are the roots of $az^2+bz+c=0$

$\Rightarrow az^2_{1}+bz_{1}+c=0$ and $az_{1}+b+c\bar{z_{1}} = 0,$ bcz $z_{1} = \frac{1}{\bar{z_{1}}}$$

$\Rightarrow az_{1}z_{2}+bz_{2}+c\bar{z_{1}}z_{2}\cdots \cdots (1)$

$\Rightarrow az^2_{2}+bz_{2}+c=0$ and $az_{2}+b+c\bar{z_{2}} = 0,$ bcz $z_{2} = \frac{1}{\bar{z_{2}}}$$

$\Rightarrow az_{1}z_{2}+bz_{1}+cz_{1}\bar{z_{2}}\cdots \cdots (2)$

$Now adding $(1)$ and $(2)$ eqn $2a(z_{1}z_{2})+b(z_{1}+z_{2})+c(z_{1}\bar{z_{2}}+\bar{z_{1}}z_{2}) =0$

$\Rightarrow 2a \cdot \frac{c}{a}-b\cdot \frac{b}{a}+c(-1) = 0 \Rightarrow b^2=ac$

He-Mann · Dec 23, 2016

Cool result. Is there a discrimnant for complex-valued quadratics?

pikachu975 · Dec 23, 2016

juantheron said:
$Let $z_{1},z_{2}$ be two roots of $az^2+bz+c=0.$ So $z_{1}+z_{2} = -\frac{b}{a}\;,z_{1}\cdot z_{2} = \frac{c}{a}$

$Given $|z_{1}| = |z_{2}| = 1$ and $|a| = |b| = |c| = 1$

$So $|z_{1}+z_{2}|^2 = 1\Rightarrow (z_{1}+z_{2})\cdot (\bar{z_{1}}+\bar{z_{2}}) = 1\Rightarrow z_{1}\bar{z_{2}}+\bar{z_{1}}z_{2} = -1$

$Now since $z_{1},z_{2}$ are the roots of $az^2+bz+c=0$

$\Rightarrow az^2_{1}+bz_{1}+c=0$ and $az_{1}+b+cz_{1} = 0,$ bcz $z_{1} = \frac{1}{\bar{z_{1}}}$$

$Solving these two equations $2a(z_{1}z_{2})+b(z_{1}+z_{2})+c(z_{1}\bar{z_{2}}+\bar{z_{1}}z_{2}) =0$

$\Rightarrow 2a \cdot \frac{c}{a}-b\cdot \frac{b}{a}+c(-1) = 0 \Rightarrow b^2=ac$

How'd you get the equation in the second last line?

juantheron · Dec 23, 2016

Sorry pikachu97 , i have edited my post.

complex number (1 Viewer)

juantheron

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calamebe

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He-Mann

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juantheron

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