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Sagot :
Using the quadratic formula, we solve for [tex]z^2[/tex].
[tex]z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2[/tex]
Taking square roots on both sides, we end up with
[tex]z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}[/tex]
Compute the square roots of -171 + 140i.
[tex]|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221[/tex]
[tex]\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)[/tex]
By de Moivre's theorem,
[tex]\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[/tex]
and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find
[tex]t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}[/tex]
as well as the fact that
[tex]0<\tan(t)<1 \implies 0along with the half-angle identities to find
[tex]\cos\left(\dfrac t2\right) = \sqrt{\dfrac{1+\cos(t)}2} = \dfrac{14}{\sqrt{221}}[/tex]
[tex]\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}[/tex]
(whose signs are positive because of the domain of [tex]\frac t2[/tex]).
This leaves us with
[tex]z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}[/tex]
Compute the square roots of 5 + 12i.
[tex]|5 + 12i| = \sqrt{5^2 + 12^2} = 13[/tex]
[tex]\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)[/tex]
By de Moivre,
[tex]\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[/tex]
and its negative, -3 - 2i. We use similar reasoning as before:
[tex]t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}[/tex]
[tex]1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4[/tex]
[tex]\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}[/tex]
[tex]\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}[/tex]
Lastly, compute the roots of -2i.
[tex]|-2i| = 2[/tex]
[tex]\arg(-2i) = -\dfrac\pi2[/tex]
[tex]\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i[/tex]
as well as -1 + i.
So our simplified solutions to the quartic are
[tex]\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}[/tex]
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