Solving A Cubic Equation in Two Ways | Problem 244

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[@scottleung9587]

I first expressed the solutions in polar form, then translated them into rectangular coordinates via the unit circle.

[@honestadministrator]

a + i b = exp ( i 2 n π /3) With n = -1, 0,1 Hereby a = cos ( 2 r π /3) b = sin ( 2 r π /3) with r = 0, 1, -1

[@honestadministrator]

a + i b = e ^ ( 2 π n i) /3 Here in n= -1, 0, 1 ( a, b) = (cos ( 2 π /3) - i sin (2 π /3)) , 1, cos ( 2 π /3) + i sin (2 π /3)) ,

[@AbouTaim-Lille]

Use the polar system. That is easier. Z= a+b.i = ρ.cis Θ So z³ = (ρ.cis Θ)³ = ρ³.cis 3Θ = 1 = 1.cis 0. So ρ = 1. And cis 3Θ = cis 0 ==> 3Θ. = 2πk. Where k is an integer. And thus Θ = ⅔ π k. So by giving the values k=0,1,2 we have 3 solutions as the other values are just a repitition. And they're Θ= 0. And Θ= ⅔π. And Θ = 4/3 π.

[@yuryp6975]

I startted wirh rhe second method but then just factored Z^3-1=0 to get the 3 solutions

[@roger7341]

Let me guess. 1 lies on the intersection of a unit circle centered on the origin of the complex plane with the real axis. (-1/2+i√3/2) lies 1/3rd of the way around that unit circle, so (-1/2+i√3/2)^3 rotates that point 240° back to 1. Thus a=-1/2 and b=√3/2. Also (-1/2-i√3/2) lies 2/3rd of the way around that unit circle, so (-1/2-i√3/2)^3 rotates that point 480° back to 1. Thus a=-1/2 and b=-√3/2.

[@stevemonkey6666]

Method 2 is more awesome!😅

[@aplusbi]

Glad you think so!