Author Topic: square roots (Read 3024 times) Tweet Share

fredrick · « **on:** October 30, 2008, 03:47:07 pm »

find the square roots of $\sqrt{21i-7}$

thanks

Glockmeister · « **Reply #1 on:** October 30, 2008, 04:20:45 pm »

Ok, I think you would need to use De Movire's Theorem for this one.

Mao · « **Reply #2 on:** October 30, 2008, 04:52:19 pm »

it'll be hard to use de moivre's, as there it will not give an exact angle [unless you really want to deal with arctan(3)]

$\mbox{Let }z=x+yi$

$(x+yi)^2 = 21i - 7$

$(x^2 - y^2)+2xyi = -7 + 21i$

$\implies 2xy = 21 \implies y = \frac{21}{2x}$

$\implies x^2 - \frac{441}{4x^2} = -7$

$\implies \frac{441}{4x^2} = x^2 + 7$

$\implies 4(x^2)^2 + 28(x^2) - 441 = 0$

$\implies x^2 = \frac{-28 + \sqrt{28^2 + 4\times 4\times 441}}{8} = \frac{14\sqrt{10}-14}{4}$ (notice that the negative option has been discarded as $x^2\ge 0$

$\implies x = \pm \frac{\sqrt{14(\sqrt{10}-1)}}{2}$

$\implies y = \pm \dfrac{21}{\sqrt{14(\sqrt{10}-1)}} = \pm \frac{3\sqrt{14(\sqrt{10}-1)}}{2(\sqrt{10}-1)} = \pm \frac{3\sqrt{14(\sqrt{10}-1)}\cdot (\sqrt{10}+1)}{2(\sqrt{10}-1)(\sqrt{10}+1)} = \pm \frac{\sqrt{14(\sqrt{10}-1)(\sqrt{10}+1)^2}}{6} = \pm \frac{3\sqrt{14(\sqrt{10}+1)}}{6} = \pm \frac{\sqrt{14(\sqrt{10}+1)}}{2}$

$\Therefore z=\left\{ \frac{\sqrt{14(\sqrt{10}-1)}}{2} + \frac{\sqrt{14(\sqrt{10}+1)}}{2}i, -\frac{\sqrt{14(\sqrt{10}-1)}}{2} - \frac{\sqrt{14(\sqrt{10}+1)}}{2}i \right\}$

FFFFFFFFFFUUUUUUUUUUUUUUUU

Damo17 · « **Reply #3 on:** October 30, 2008, 04:55:23 pm »

Quote from: Mao on October 30, 2008, 04:52:19 pm

it'll be hard to use de moivre's, as there it will not give an exact angle [unless you really want to deal with arctan(3)]

$\mbox{Let }z=x+yi$

$(x+yi)^2 = 21i - 7$

$(x^2 - y^2)+2xyi = -7 + 21i$

$\implies 2xy = 21 \implies y = \frac{21}{2x}$

$\implies x^2 - \frac{441}{4x^2} = -7$

$\implies \frac{441}{4x^2} = x^2 + 7$

$\implies 4(x^2)^2 + 28(x^2) - 441 = 0$

$\implies x^2 = \frac{-28 + \sqrt{28^2 + 4\times 4\times 441}}{8} = \frac{14\sqrt{10}-14}{4}$ (notice that the negative option has been discarded as $x^2\ge 0$

$\implies x = \pm \frac{\sqrt{14(\sqrt{10}-1)}}{2}$

$\implies y = \pm \dfrac{21}{\sqrt{14(\sqrt{10}-1)}} = \pm \frac{3\sqrt{14(\sqrt{10}-1)}}{2(\sqrt{10}-1)} = \pm \frac{3\sqrt{14(\sqrt{10}-1)}\cdot (\sqrt{10}+1)}{2(\sqrt{10}-1)(\sqrt{10}+1)} = \pm \frac{\sqrt{14(\sqrt{10}-1)(\sqrt{10}+1)^2}}{6} = \pm \frac{3\sqrt{14(\sqrt{10}+1)}}{6} = \pm \frac{\sqrt{14(\sqrt{10}+1)}}{2}$

$\Therefore z=\left\{ \frac{\sqrt{14(\sqrt{10}-1)}}{2} + \frac{\sqrt{14(\sqrt{10}+1)}}{2}i, -\frac{\sqrt{14(\sqrt{10}-1)}}{2} - \frac{\sqrt{14(\sqrt{10}+1)}}{2}i \right\}$

FFFFFFFFFFUUUUUUUUUUUUUUUU

WOW!!!

shinny · « **Reply #4 on:** October 30, 2008, 04:56:24 pm »

EDIT: Oh whoops that's what you did. Thought you tried how I interpreted and found the square roots of the root of 21i-7 =T Didn't think that question would be so hard o_O

vce08 · « **Reply #5 on:** October 30, 2008, 04:57:11 pm »

I assume this won't pop up on the VCAA paper?

Mao · « **Reply #6 on:** October 30, 2008, 04:59:37 pm »

i certainly hope not!

fredrick · « **Reply #7 on:** October 30, 2008, 05:04:39 pm »

Thanks Mao i got half way and ended up saying the same word as you did FFFFFFFFFFUUUUUUUUUUUUUUUU

Glockmeister · « **Reply #8 on:** October 30, 2008, 05:16:46 pm »

yeah... i tried De Movire and i was looking at the $tan^{-1} (3)$ thing and I was going fuck this. So I didn't finish.

Mao · « **Reply #9 on:** October 30, 2008, 05:24:54 pm »

method 2 [I'm not crazy I'm just a little unwell, I know, coz right now you can't tell.... but stay a while maybe then you'll see, what English does to me~]:

$\mbox{Let }w=-7+21i = (7\sqrt{10})\mbox{cis}(\tan^{-1}(-3)+\pi)$

$\mbox{Let }z^2 = w$

$\implies z = \sqrt{7\sqrt{10}}\mbox{cis}\left(\frac{1}{2}(\tan^{-1}(-3)+\pi)\right)$

$\cos(\tan^{-1}x) = \frac{1}{\sqrt{x^2+1}}\; \sin(\tan^{-1}x)=\frac{x}{\sqrt{x^2+1}}$

$\implies \cos(\tan^{-1}-3) = \frac{1}{\sqrt{10}}\; \sin(\tan^{-1}-3)=\frac{-3}{\sqrt{10}}$

now, $\cos\left(\frac{x}{2}\right) = \frac{\sqrt{2\cos(x)+2}}{2}$ (from double angle formula),

hence $\cos \left(\frac{1}{2}(\tan^{-1}(-3)+\pi)\right) = \frac{\sqrt{2\cos(\tan^{-1}(-3)+\pi)+2}}{2}=\frac{\sqrt{2\cdot\left( \frac{1}{\sqrt{10}}\cdot -1 - 0\right)+2}}{2}=\frac{\sqrt{\frac{2\sqrt{10}-2}{\sqrt{10}}}}{2}=\sqrt{\frac{\sqrt{10}-1}{2\sqrt{10}}}$

and, $\sin\left(\frac{x}{2}\right) = \frac{\sqrt{2-2\cos(x)}}{2}$

hence $\sin \left(\frac{1}{2}(\tan^{-1}(-3)+\pi)\right) = \frac{\sqrt{2-2\cos(\tan^{-1}(-3)+\pi)}}{2}=\frac{\sqrt{2-2\cdot\left( -\frac{1}{\sqrt{10}}\right)}}{2}=\frac{\sqrt{\frac{2\sqrt{10}+2}{\sqrt{10}}}}{2}=\sqrt{\frac{\sqrt{10}+1}{2\sqrt{10}}}$

hence, $\sqrt{7\sqrt{10}}\sin \left(\frac{1}{2}(\tan^{-1}(-3)+\pi)\right) = \sqrt{\frac{7(\sqrt{10}+1)}{2}}=\frac{\sqrt{14(\sqrt{10}+1)}}{2}$

and $\sqrt{7\sqrt{10}}\cos \left(\frac{1}{2}(\tan^{-1}(-3)+\pi)\right) = \sqrt{\frac{7(\sqrt{10}-1)}{2}}=\frac{\sqrt{14(\sqrt{10}-1)}}{2}$

so a solution is $\frac{\sqrt{14(\sqrt{10}-1)}}{2} + \frac{\sqrt{14(\sqrt{10}+1)}}{2}i$

now, since the two solutions are $\pi$ apart, we then multiply $\mbox{cis}(\pi) = -1$

$\Therefore z=\left\{ \frac{\sqrt{14(\sqrt{10}-1)}}{2} + \frac{\sqrt{14(\sqrt{10}+1)}}{2}i, -\frac{\sqrt{14(\sqrt{10}-1)}}{2} - \frac{\sqrt{14(\sqrt{10}+1)}}{2}i \right\}$

Glockmeister: here you go. this method is fucking insane

in hindsight, this method required less competency with radicals and fractions and etc, though the previous method is conceptually easier.

fredrick · « **Reply #10 on:** October 30, 2008, 05:26:23 pm »

Just one more, in a similar question(not as ugly as the other one) it says find the square roots of $\sqrt{24i-7}$ .

I ended up with $3+4i$ and $-3-4i$ , however the ansers proceed to find the square roots of $3+4i$ and $-3-4i$

EDIT: wow nice method 2!

fredrick · « **Reply #11 on:** October 30, 2008, 05:47:46 pm »

Thanks! makes sence now, i knew it required 4 solutions

Mao · « **Reply #12 on:** October 30, 2008, 05:52:14 pm »

HOMEWORK EXERCISE FOR THOSE WANTING TO DIE TONIGHT:

find the quartic roots of $21i-7$

Glockmeister · « **Reply #13 on:** October 30, 2008, 05:53:04 pm »

Fuck this, I've got an English exam tomorrow

fredrick · « **Reply #14 on:** October 30, 2008, 06:15:59 pm »

Do you mean quartic roots of $\sqrt{21i-7}$

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