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Question Number 95062 by mr W last updated on 22/May/20

$$solveforx,y,z\in \mathbb{C}suchthat$$$$\mid x\mid =\mid y\mid =\mid z\mid =1$$$$x+y+z=1$$$$xyz=1$$

Commented by EmericGent last updated on 22/May/20

BlackPenRedPen solved it today ^^ (1;i;-i)

Commented by mr W last updated on 23/May/20

$$maybeotherpeoplehaveothermethods.$$$$basicallyalmostallexistingquestions$$$$intheworldhavealreadysolutions.$$$$ifonlyquestions,whicharenotyet$$$$solved,areinterestingforthisforum,$$$$thenwecancloseitrightnow.$$

Commented by EmericGent last updated on 22/May/20

I know, but posted the video this morning

Commented by mr W last updated on 22/May/20

$$doyouhaveasolution?i^{\prime }m$$$$interested.interestingisitwhen$$$$differentpeopletakedifferentways$$$$tosolvethesameproblem.$$

Commented by prakash jain last updated on 23/May/20

$$x=e^{ia},y=e^{ib},z=d^{ic}$$$$e^{i(a+b+c)}=1$$$$\cos (a+b+c)=1$$$$\sin (a+b+c)=0$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=2n\pi$$$$x+y+z=1$$$$\cos \mathrm{a}+\cos \mathrm{b}+\cos \mathrm{c}=1$$$$\sin \mathrm{a}+\sin \mathrm{b}+\sin \mathrm{c}=0$$$$\cos \mathrm{a}+\cos \mathrm{b}+\cos (\mathrm{a}+\mathrm{b})=1$$$$\cos \mathrm{a}+\cos \mathrm{b}-1=\cos (\mathrm{a}+\mathrm{b})\left(\mathrm{i}\right)$$$$\sin \mathrm{a}+\mathrm{sinb}-\sin (\mathrm{a}+\mathrm{b})=0$$$$\sin \mathrm{a}+\sin \mathrm{b}=\sin (\mathrm{a}+\mathrm{b})\left(\mathrm{ii}\right)$$$${\left(\mathrm{i}\right)}^2+{\left(\mathrm{ii}\right)}^2$$$${\cos }^2\mathrm{a}+{\cos }^2\mathrm{b}+1+{\sin }^2\mathrm{a}+{\sin }^2\mathrm{b}$$$$+2\sin \mathrm{asin}\mathrm{b}+2\cos \mathrm{acos}\mathrm{b}$$$$-2\cos \mathrm{a}-2\cos \mathrm{b}=1$$$$1+\cos (\mathrm{a}-\mathrm{b})=\cos \mathrm{a}+\cos \mathrm{b}$$$$\cos (\mathrm{a}-\mathrm{b})=\mathrm{cosa}+\mathrm{cosb}-1$$$$\cos (\mathrm{a}-\mathrm{b})-\cos (\mathrm{a}+\mathrm{b})=0$$$$\cos a\cos b=0\left(\mathrm{II}\right)$$$$\mathrm{We}\mathrm{can}\mathrm{take}$$$$\cos \mathrm{a}=0$$$$a=2n\pi \pm \frac{\pi }{2}$$$$\mathrm{solve}\mathrm{for}\mathrm{one}\mathrm{value}$$$$x=i$$$$y+z=1-i$$$$yz=-i$$$${(y+z)}^2-4yz=-2i+4i$$$${(y-z)}^2=2i={(1+i)}^2$$$$y-z=1+i$$$$y=1$$$$z=-i$$$$\mathrm{It}\mathrm{can}\mathrm{seen}\mathrm{from}\mathrm{symettry}\mathrm{that}$$$$\mathrm{alternate}\mathrm{solution}\mathrm{for}\left(\mathrm{II}\right)\mathrm{and}$$$$\mathrm{subsequent}\mathrm{equation}\mathrm{only}$$$$\mathrm{interchanges}x,y,z$$

Commented by mr W last updated on 23/May/20

$$verynice!thankssir!$$

Commented by mr W last updated on 23/May/20

$$x,y,zarerootsofeqn.$$$$t^3-t^2+kt-1=0$$$$\mid t\mid =1$$$$itmusthaveatleastonerealroot,$$$$thisrealrootmustbe-1or1.$$$$-1-1-k-1=0\Rightarrow k=-3$$$$1-1+k-1=0\Rightarrow k=1$$$$$$$$withk=-3:$$$$t^3-t^2-3t-1=0$$$$(t+1)(t^2-2t-1)=0$$$$\Rightarrow t=-1,1\pm \sqrt{5}$$$$but\mid 1\pm \sqrt{5}\mid \ne 1,\Rightarrow k\ne -3,t\ne -1$$$$$$$$withk=1:$$$$t^3-t^2+t-1=0$$$$(t-1)(t^2+1)=0$$$$\Rightarrow t=1,\pm i$$$$\Rightarrow x,y,z\in (1,i,-i)$$

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