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Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

This is a way to work it out

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Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$