Author Topic: square roots (Read 3017 times) Tweet Share

Mao · « **Reply #15 on:** October 30, 2008, 06:24:28 pm »

um..... no....

a square root of $21i - 7$ is $\sqrt{21i-7}$ (i forgot to point that out to you)

so square roots of $\sqrt{21i-7}$ is actually quartic roots of $21i-7$ [and if that's what you actually mean, GET THE FUCK OFF HERE NAO]

fredrick · « **Reply #16 on:** October 30, 2008, 06:45:37 pm »

OK now im confused...didnt we(you) just work out the quartic roots of $21i-7$

shinny · « **Reply #17 on:** October 30, 2008, 07:03:43 pm »

No...Mao found the square roots of 21i-7, not quartic. That's what my post was about before because I was wondering whether you meant the square roots of $\sqrt{21i-7}$ (aka. 2 of the quartic roots of 21i-7) or just 21i-7.

dcc · « **Reply #18 on:** October 30, 2008, 07:04:00 pm »

Quote from: Mao on October 30, 2008, 05:52:14 pm

find the quartic roots of $21i-7$

$z^4 = r^4 \ \mbox{cis} \left( 4 \theta \right) = 21i - 7 = 7 \sqrt{10} \ \mbox{cis} \left( \pi - \tan^{-1} \left( 3 \right) \right)$

$r^4 = 7 \sqrt{10} \implies r = \sqrt[4]{7} \cdot \sqrt[8]{10}$

$4\theta = \pi - \tan^{-1} \left(3\right),\ 3\pi - \tan^{-1} \left(3\right),\ 5\pi - \tan^{-1} \left(3\right),\ 7\pi - \tan^{-1} \left(3\right)$

Therefore our four roots are:

$z_{1} = \sqrt[4]{7} \cdot \sqrt[8]{10} \ \mbox{cis} \left( \dfrac{\pi - \tan^{-1} \left(3\right)}{4} \right)$

$z_{2} = \sqrt[4]{7} \cdot \sqrt[8]{10} \ \mbox{cis} \left( \dfrac{3\pi - \tan^{-1} \left(3\right)}{4} \right)$

$z_{3} = \sqrt[4]{7} \cdot \sqrt[8]{10} \ \mbox{cis} \left( \dfrac{5\pi - \tan^{-1} \left(3\right)}{4} \right)$

$z_{4} = \sqrt[4]{7} \cdot \sqrt[8]{10} \ \mbox{cis} \left( \dfrac{7\pi - \tan^{-1} \left(3\right)}{4} \right)$

(If VCAA set a question anything similar to finding the nth roots of something, it would be NICE multiples of the cosine/sine of the basic angles

)

p.s. ceebz simplifying

bturville · « **Reply #19 on:** October 30, 2008, 07:05:10 pm »

TOO MUCH MATHS!

fredrick · « **Reply #20 on:** October 30, 2008, 07:31:03 pm »

yeh sorry the first post i meant was to find the quartic roots of it...nvm confusion sorted

Mao · « **Reply #21 on:** October 30, 2008, 07:52:59 pm »

ARE YOU FUCKING SERIOUS?!

shinny · « **Reply #22 on:** October 30, 2008, 07:55:22 pm »

YA HES SRS.

Go Mao Go. Cartesian solution here we come.

Mao · « **Reply #23 on:** October 30, 2008, 08:54:10 pm »

fuck that, I'm not writing it out

but here is the general form:

$\sqrt{a+bi} = \frac{\sqrt{2\sqrt{a^2+b^2}+2a}}{4} + \mbox{sng}(b)\frac{\sqrt{2\sqrt{a^2+b^2}-2a}}{4}i$

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